o

We consider semilinear obstacle problem with measure data associated with uniformly elliptic divergence form operator. We prove existence and uniqueness of entropy solution of the problem and provide stochastic representation of the solution in terms of some generalized reflected backward stochastic differential equations with random terminal time.


Introduction
Let D ⊂ R d (d ≥ 2) be a bounded domain with regular boundary.In this paper we investigate obstacle problem with measure data associated with semilinear operator of the form and a : D → R d ⊗ R d is a measurable symmetric matrix-valued function such that for some Λ ≥ 1, f : D × R → R is a measurable function satisfying some assumptions to be specified later on. ( The measure γ is uniquely determined by (1.5), and is called the obstacle reaction associated with u.In the general case where µ ∈ M 2 b (D) we consider entropy solutions of (1.3) in the sense defined in [13], i.e. we call u : D → R a solution of (1.3) if there exists a positive measure γ ∈ M 2 b (D) such that u is a quasi-continuous entropy solution of the problem Au = −µ − γ, u| ∂D = 0, u ≥ ψ q.e.
In the analytical part of the paper we show that under (H1)-(H3) the obstacle problem (1.3) has a unique entropy solution.In the proof of that part we combine ideas from [13], where the obstacle problem with f = 0 but more general than A nonlinear elliptic operator of monotone type Ā mapping W 1,p (D), p > 1, into its dual is considered, and from [1,2], where problems Āu + f u = −µ, u| ∂D = 0 (i.e.ψ = −∞) with f satisfying (H1) are considered.It is known that if µ ∈ L p (D) with p > d and f satisfies the Lipschitz and the linear growth condition in u then the Dirichlet problem (1.6) with ψ = −∞ has a unique continuous weak solution which can be represented by solutions of some backward stochastic differential equation (BSDE) with random terminal time (see [16,17]).It is also known that viscosity solutions of some problems of the form (1.6) with nondivergence form operator in place of A may be represented by solutions of some reflected BSDEs (RBSDEs) with random terminal time (see [15]).Therefore it is natural to try to relate solutions of (1.6) to some reflected BSDE with forward driving process associated with A. In the paper we show that this is indeed possible and leads to investigation of interesting generalized RBSDEs involving additive functionals associated with measures µ and γ.
Let X = {(X, P x ); x ∈ R d } be a Markov process associated with the operator A (see Section 2) and let X D be the part of X on D, i.e.X D = {(X D , P x ); x ∈ D ∪ {∂}}, where ∂ is an extra point adjoint to D, and τ = inf{t ≥ 0 : X t ∈ D}.
It is known that to every µ ∈ M 2 b (D) corresponds a unique continuous additive functional (CAF) R of X D whose Revuz measure is µ.The main result of the paper says that if (H1)-(H3) are satisfied then there exists a unique solution to (1.6) which has a quasicontinuous and quasi-everywhere (q.e. for short) finite version u.Secondly, for q.e.
x ∈ D the triple (Y, Z, K), where K is a positive CAF of X D in Revuz correspondence with γ and Y t = u(X D t ), Z t = σ∇u(X D t ), t ≥ 0, where σ is a symmetric square root of a, is a unique solution to the generalized reflected BSDE of the form t∧τ Z s dB s , t ≥ 0, P x -a.s., Y t ≥ ψ(X D t ), P x -a.s. for t ≥ 0, K is a continuous increasing, K 0 = 0, τ 0 (Y s − ψ(X s )) dK s = 0, P x -a.s., where B is a Wiener process.It follows immediately that for q.e.x ∈ D, u(x) = Y 0 , P x -a.s..
(1.9) Thus, the above RBSDE provides stochastic representation of quasi-continuous solutions of (1.6).With this representation in hand the minimality of u in the sense defined in [13] is a consequence of comparison results for solutions of generalized BSDEs proved in Section 3. From (1.8) and the fact that K increases only when Y = ψ(X D ) we also deduce that D (u − ψ) dγ = 0, i.e. the obstacle reaction γ associated with u is minimal in the sense that it acts only when u = ψ.Finally, let us mention that the representation (1.9) makes it posssible to give simple probabilistic definition of a solution of the problem (1.6).
Notation.As usual, for p ∈ [1, ∞] we denote by L p (D) and W 1,p (D) the standard Lebesgue and Sobolev spaces, is the dual space to H 1 0 (D).By • 2 and (• , •) 2 we denote the usual norm and scalar product in L 2 (D).

Additive functionals of symmetric diffusions and smooth measures
In this section we are concerned with additive functionals of killed symmetric diffusions associated with A which are in the Revuz correspondence with smooth measures on D.

Symmetric diffusions
Let Ω = C([0, ∞), R d ) denote the space of continuous R d -valued functions on [0, ∞) equipped with the topology of uniform convergence on compact intervals and let X be the canonical process on Ω.It is known that given operator A defined by (1.1) with a satisfying (1.2) one can construct a weak fundamental solution p for A and then a time-homogeneous Markov process X = {(X, P x ); x ∈ R d } for which p is the transition density function, i.e.
for any Borel B ⊂ R d (see, e.g., [20]).Alternatively, one can define X as the Markov process associated with the Dirichlet form ∞ denote the completion of F 0 ∞ with respect to P µ and let F µ t (resp.FT,µ t ) denote the completion of F 0 t (resp.FT,0 Let X D denote the process X killed outside D, i.e.X D is defined by (1.7), where ∂ is an isolated point regarded as the point at infinity of D, and let X D denote the part of X on D, i.e.X D = {(X, P x ), x ∈ D ∪ {∂}}, where P ∂ (X whenever the limit exists.A CAF A of X such that e(A) = 0 and for t > 0, We say that a CAF A of X in the strict sense is of zero quadratic variation if for each for any T > 0 and any sequence Let us recall that from [16,Theorem 3.4] it follows that there exist a continuous MAF M of X in the strict sense and a CAF A of X in the strict sense of zero quadratic variation such that Thus, for each x ∈ R d the canonical process X is an ({F t }, P x )-Dirichlet process in the sense of Föllmer.Note also that the decomposition (2.1) coincides with the Fukushima strict decomposition of X into a MAF of X of locally zero energy and a CAF of X of zero energy (see [8,Theorem 5.5.1]).
From [16,Theorem 3.4] one can conclude that for every T > 0 there exists a unique continuous { FT t } -adapted process N T such that N T is a square-integrable martingale on [0, T ] under P x for each x ∈ R d and where (here D j p stands for the generalized derivative of y j → p(t, x, y)).Moreover, the covariation processes of M = (M 1 , . . ., M d ) and N T = (N T,1 , . . ., N T,d ) are given by (2.4) 3) may be called the strict Lyons-Zheng decomposition of X.

Capacity, smooth measures
Let F be a compact subset of D. Recall that the capacity of F with respect to D is defined as Finally, the capacity of any B ⊂ D is defined as By [8, Theorem 4.2.1(ii)],N ⊂ D is exceptional iff cap(N ) = 0. Hence, in particular, for any Borel B ⊂ D, cap(B) = 0 iff P m (∃t > 0, X D t ∈ B) = 0.In what follows a statement depending on x ∈ B is said to hold quasieverywhere on B (q.e. for short) if there is a set N ⊂ B of zero capacity such that the statement holds for every B \ N .
A function u : D → R is quasi-continuous if for every ε > 0 there is an open set E such that cap(E) < ε and u| D\E is continuous in D \ E.
It is known that every u ∈ H 1 0 (D) has a quasi-continuous representative that will always be identified with u.
Let M(D) denote the set of all signed Radon measures on D and let M b (D) denote the subset of M(D) consisting of all measures whose total variation |µ| on D is finite.As usual, we identify M b (D) with the dual of the Banach space C 0 (D) of continuous functions on D which vanish on the boundary of D, so that the duality is given by µ, u where now •, • denotes the duality pairing between H −1 (D) and H 1 0 (D) and u on the right hand-side is a quasi-continuous representative of u on the left hand-side.
Finally, let us recall that in [3] the following important result is proved:

Additive functionals of killed diffusions and smooth measures
In this subsection we use decomposition (2.6) to investigate structure of additive functionals of X D corresponding to measures of the class M 2 b (D).Let S + 0 (D) denote the family of positive Radon measures on D of finite energy integrals, i.e. such that for some C ≥ 0 (C 0 (D) is the space of all continuous functions on D having compact support).It is known (see [8,Section 2.2]) that µ ∈ S + 0 (D) iff for each α > 0 there exists a unique function (2.7) By [8,Theorem A.2.10] the part process X D is a Markov process on D (with respect to the filtration {F t }) with the transition function Therefore the semigroup {P D t } of operators associated with X D is given by where E x denotes the expectation with respect to P x and B + (D) is the space of positive measurable functions on D. By [5, Theorem 2.4], p D (t, x, •) admits the transition density p D (t, x, y), which is symmetric and continuous on D × D.
From now on we will use the following useful convention: any numerical function f on D will automatically be extended to D ∪ {∂} by setting f (x) = 0, x ∈ ∂D, f (∂) = 0.
and let Definition.We say that a PCAF A of X D and µ ∈ M 2,+ b (D) are in the Revuz correspondence if for any α > 0 and f, g In that case we call µ the Revuz measure of A and we write µ ∼ A or A ∼ µ.
It is known (see [8,Theorem 5.1.3])that (2.8) is equivalent to the following condition: for any t > 0, g, f ∈ B + (D), (2.9) By Lemma 5.1.8and Theorem 5.1.3in [8], any µ ∈ M 2,+ b (D) admits a PCAF A of X D whose Revuz measure is µ and that the PCAF A related to given µ ∈ M 2,+ b (D) is unique up to the equivalence.In particular, if µ(dx) = f (x) dx for some positive f ∈ L 1 (D) then the unique PCAF A of X D associated with µ is given by Given a signed measure µ ∈ S 0 we decompose it as µ = µ 1 − µ 2 in the above way and set Clearly A is a finite CAF of bounded variation and does not depend on the choice of µ 1 , µ 2 .

Lemma 2.2.
If A is a CAF of X D of finite variation associated with some measure µ ∈ M 2,+ b (D) then E ν A τ < ∞ for every ν ∈ S 00 (D).Proof.By (2.11), for any ν ∈ S 00 (D) and N > 0 we have We only consider the case T = ∞.The proof of the lemma in case T ∈ [0, ∞) is similar and therefore we omit it.For any ε > 0 i N > 0 we have from which the result follows.
Following [18] given T > 0 and h = (h 1 , . . ., h d ) ∈ L p (D) d with p > d we set where M, V, N T are processes of the decomposition (2.2).One can show (see [18]) that In the sequel we will use the notation where a −1 denotes the inverse of a. From the above and (2.10) it follows that if h ∈ W 1,p (D) d for some p > d then H(a −1 h) is a CAF of X D in the strict sense.Applying approximation arguments one can show that in fact it is a CAF of X D in the strict sense for any h ∈ L p (D) d with p > d.
The following proposition is a variant of [10, Proposition 3.5].

Then
(i) There is a subsequence (still denoted by n) and a CAF A of X D such that for every T > 0, for a.e.x ∈ D. (2.12) First we are going to show that for every T > 0, where By the definition of A n , By Doob's L 2 -inequality and symmetry of the transition density p D (t, •, •), ( Furthermore, From (2.13) it follows that there is a subsequence such that if n, k → ∞ along this subsequence then E x (Y n,k T |G σ ) → 0, P x -a.s. for a.e.x ∈ D. Hence P x (σ < ∞) = 0 for a.e.x ∈ D, and so P m (σ < ∞) = 0. From (2.5) we conclude now that cap(F ) = 0, hence that cap(B) = 0. Thus, lim n,k→∞ for q.e.x ∈ D. Hence for q.e.x ∈ D there exists a continuous process A x such that To complete the proof of (i) we use arguments from the proof of [8,Lemma A.3.2].Set for q.e.x ∈ D, applying the Borel-Cantelli lemma shows that P x (Λ) = 1 for q.e.x ∈ D.
(ii) Without loss of generality we may and will assume that µ ≥ 0. Let j n be a mollifier and let µ n = µ * j n .Then µ n = divg n , where g n = g * j n .Since µ n 1 ≤ µ M b (D) , {µ n } is relatively compact in the weak * topology in M b (D) by Alaoglu's theorem.Therefore choosing a subsequence if necessary we may assume that µ n µ weakly * in M b (D).Let A n be the AF defined by (2.12).Since A n ∼ µ n , for every f, g ∈ B(D) and α > 0, On the other hand, by part (i), Thus, . By [8, Lemma 5.1.7]there exists a smooth measure µ A ∈ M 2,+ (D) such that µ A ∼ A. Since (2.17) is satisfied with µ replaced by µ A , repeating arguments from the proof of [8,Theorem 5.1.3]shows that f • µ A , g = f • µ, g for any 0-excessive function h.Since for every x ∈ ∂D, Proof.By Riesz's theorem there exist u, On the other hand, putting then there is a subsequence (still denoted by n) such that for any ν ∈ S + 00 (D) and T > 0, Proof.In view of Lemmas 2.3 and 2.5 it suffices to prove the proposition in the case for a.e.x ∈ D, and therefore in much the same way as in the proof of (2.13) one can show that for any T > 0, E m sup t≤T |A n t − A t | → 0. To prove the lemma it suffices now to repeat arguments from the proof of (2.16) to show that up to a subsequence, sup t≤T |A n t∧τ − A t∧τ | → 0 in measure P x for q.e.x ∈ D, and hence, by (2.7), in measure P ν for any ν ∈ S + 00 (D).

RBSDEs and the obstacle problem -uniqueness of solutions
Let σ denote the symmetric square-root of a. Set and observe that from (2.4) it follows that B is an ({F t }, P x )-standard Brownian motion for each x ∈ R d .
Definition We say that a triple (Y Elliptic obstacle problems with measure data (v) K x is a continuous increasing process such that K x satisfy (ii), (iii) and condition (i) is satisfied with K x ≡ 0.
For a given constant k > 0 we define the truncature operator T k : R → R as and for a function u : R → R we define the truncated function T k u pointwise, i.e.
We say that a measurable and almost everywhere finite function u : D → R is an entropy solution of the problem Following [13] we adopt the following definition.
Definition We say that u : D → R is an entropy solution of OP(f, µ, ψ) if such that v ≥ ψ q.e. in D, then v ≥ u q.e. in D.
Let us remark that by the definition, if there exists a solution to OP(f, µ, ψ) then it is unique, and if u denotes the solution then the measure γ satisfying (i), (ii) is uniquely determined.We call γ the obstacle reaction associated with u.
Since entropy solution u to OP(f, µ, ψ) satisfies (3.3), it has a quasi-continuous representative.Therefore we will always assume that u denotes the quasi-continuous representative of a given entropy solution.If, in addition, ∇T k u 2 ≤ C(1 + k) for some C > 0 then the quasi-continuous representative is q.e.finite, i.e. cap{|u| = ∞} = 0 (see [7,Remark 2.11]).
We know that solution of OP(f, µ, ψ) if exists is unique by the definition.Uniqueness of solutions of associated RBSDEs with data f, µ, ψ under monotonicity condition on f follows from the following comparison result.
where L 0 (Y − Y ) denote the local time at 0 of the semimartingale Y − Y .By the assumptions on f, f , τ ∧τn t∧τ ∧τn

Existence and stochastic representation of solutions of the obstacle problem
Our main goal is to prove existence and stochastic representation of solutions of the obstacle problem with data f, µ, ψ satisfying (H1)-(H3).Since the proof of this result is rather lengthy, we first assume additionally that µ ∈ H −1 (D) and f is bounded, and then we consider the general case.
Step 1.We first assume additionally that µ ∈ L ∞ (D).Let u n ∈ H 1 0 (D) be a unique weak solution of the problem the last inequality being a consequence of (H1a) and the fact that ψ * ≥ ψ a.e.. From the above and Poincaré's inequality it follows that {u n } is bounded in H 1 0 (D).Therefore, taking a subsequence if necessary we may and will assume that there is w ∈ H 1 0 (D) such that u n → w weakly in H 1 0 (D).It is known (see [17]) that for every x ∈ D the pair By Remark 3.3, Y n ≤ Y n+1 , P x -a.s. for every x ∈ D. It follows in particular that u n ≤ u n+1 , n ∈ N, and hence that u n ≤ w a.e..As a consequence, letting n → ∞ and using the fact that f un → f w in L 2 (D) by Nemitskii's theorem (see [11,Theorem 2.1]), we see that for v ∈ K ψ , which shows that w is a solution of EVI(f, µ, ψ), i.e. w = u.Now, let R ∼ µ, K ∼ γ and let Finally, from [8, Lemma 5.1.5]and (2.7) it follows that there is a subsequence (still denoted by n) such that P ν -a.s. the sequence {u n (X)} converges to u(X) uniformly in Letting T → ∞ in (4.4), (4.5) shows that under P ν the triple defined by (4.1) is a solution of RBSDE with data f, µ, ψ and hence, by (2.7), is a solution of RBSDE x (f, µ, ψ) for q.e.
Step 2. We now show how to dispense with the assumption that µ ∈ L ∞ (D).
j n be a mollifier and let µ n = T n g 0 − div((T n g) * j n ).Let u n ∈ H 1 0 (D) be a weak solution of EVI(f, µ n , ψ) and let γ n be the obstacle reaction associated with u n so that is a solution of RBSDE x (f, µ n , ψ).Hence, for any T > 0, for q.e.x ∈ D, where R n ∼ µ n , K n ∼ γ n .Let u ∈ H 1 0 (D) be a weak solution of EVI(f, µ, ψ) and let γ be the obstacle reaction associated with u.Taking v = u n as a test function in (1.4) D) by Nemitskii's theorem, and hence γ n → γ in H −1 (D).From Lemma 2.6 and (2.7) it follows now that there is a subsequence (still denoted by n) such that for any ν ∈ S + 00 (D), where R ∼ µ, K ∼ γ.To complete the proof it suffices now to let n → ∞ in (4.Then for q.e.x ∈ D the pair Proof.Follows from the proof of Proposition 4.1 with ψ = −∞ and γ = 0, K = 0.
Proof.Let us define the operator B : Let u n ∈ H 1 0 (D) be a solution of EVI(f, µ, ψ n ) with ψ n = ψ − n −1 and let w n ∈ H 1 0 (D) be a solution of the elliptic variational inequality with the operator B, measure µ − µ * and obstacle ψ n − ψ * , i.e.

−Bw
Since Aψ * = −µ * , for every η ∈ K ψn we have From the above it follows that w n + ψ * is a solution of EVI(f, µ, ψ n ), i.e. u n = w n + ψ * .As a consequence, the obstacle reaction β n associated with w n coincides with the obstacle reaction γ n associated with u n because ) is bounded and satisfies (H1a), it follows from Proposition 4.2 and Remark 3.
By the above inequalities, Hence ξ = w n , and so we have From this it may be concluded that Indeed, since v ≥ w n , (4.8) will be proved once we prove that γ n ({v = w n }) = 0. Since v ≥ 0 and ψ n − ψ * < 0, On the other hand, by (2.9), where K n ∼ γ n .Since X has continuous trajectories, and hence that γ n ({u n > ψ n }) = 0. Thus, γ n ({v = w n }) = 0, and (4.8) is proved.To complete the proof of the proposition it suffices now to prove that γ n → γ in M b (D), where γ is the obstacle reaction associated with the solution u of EVI(f, µ, ψ).To see this, we first show that {u n } is bounded in H 1 0 (D).Since u n ≤ u n+1 by Proposition 4.1 and Theorem 3.1, it follows that there is w ∈ H 1 0 (D) such that u n → w weakly in H 1 0 (D).In fact, as in Step 1 of the proof of Proposition 4.1 one can show that w = u and u n → u strongly in H 1 0 (D).By [13, Proposition 3.8], u n is a solution of OP(0, μn , ψ n ) whereas u is a solution of OP(0, μ, ψ), where μn = be a weak solution of the problem Av n = −µ− γn , and let u n ∈ H 1 0 (D) be a solution of EVI(f, µ, ψ n ) with ψ n = ψ ∧ v n .By Propositions 4.1, 4.2 and Theorem 3.1, v n ≥ u n , so the proof will be completed by showing that v n ↑ v, u n ↑ u q.e. in D. To see this, let us first observe that by Proposition 4.2 and Remark 3.3, v n ≤ v n+1 q.e.Hence ψ n ≤ ψ n+1 , so using once again Propositions 4.1, 4.2 and Theorem 3.1 we see that u n ≤ u n+1 .By the above there are v * , u , it follows from the stability results for entropy solutions (see Theorem 2.3 and Corollary 3.2 in [13] Since the last problem has a unique solution, v = v * , and consequently, v n ↑ v q.e. in D. On the other hand, by the definition of a weak solution of EVI, hence that {u n } is bounded in H 1 0 (D).Therefore we may assume that u n → u * weakly in H 1 0 (D).In fact, since we already know that u from which it follows that u n → u * strongly in H 1 0 (D).Therefore letting n → ∞ in (4.10) shows that u * is a solution of EVI(f, µ, ψ).Accordingly u = u * , and consequently, u n ↑ u q.e. in D.

General measure data
Let M q (D), q ≥ 1, denote the Marcinkiewicz space of order q (see, e.g., [12,Section 2.18]).Recall that M q (D) can be defined as the set of measurable functions u : D → R such that the corresponding distribution function In the proof of the existence of a solution to the problem OP(f, µ, ψ) under (H1)-(H3) we will need the following stability result for entropy solutions of (3.2).
Theorem 4.5.Assume that f satisfies (H2) and µ, µ n ∈ M 2 b (D).Let u be an entropy solution of the problem (3.2) and u n be an entropy solution of the problem Proof.The proof follows closely the proof of [1, Theorem 6.1] (see also [2]).Nevertheless we provide its main ingredients because we will use them in the proof of our main result.
We first assume that d ≥ 3.By (3.4) with v = 0 and the fact that u n f un ≤ 0, It follows that {∇T k u n } n is bounded in L 2 (D) and hence, by Poincaré's inequality, in ).Since the imbedding H 1 0 (D) → L q (D) is compact, we may and will assume that {T k u n } n is a Cauchy sequence in L q (D).From this and estimates of meas{|u n −u m | > t} on pages 256-256 in [1] it follows that {u n } is a Cauchy sequence in measure.Hence there is u such that u n → u in measure in D. Extracting a subsequence if necessary we may and will assume that u n → u a.e.. Since f (x, •) is continuous, f un → f u a.e.. Let {ξ i } be a sequence of real smooth increasing functions such that ξ i (y) → ξ(y), where ξ(y) = 0 if |y| ≤ k and ξ(y Letting k ↓ 0 in (4.12) we see that f un 1 ≤ µ n M b (D) .Hence, by Fatou's lemma, . By (4.11) and [1, Lemma 4.2], for every ε > 0 there exists A > 0 such that meas{|∇u n | > A} ≤ ε for all n.Moreover, Using the above estimate one can show as in [1] (see pages 257-258) that {∇u n } is a Cauchy sequence in measure.Hence {∇u n } converges in measure to some function v. Since we know that for each k > 0, {∇T k u n } n is bounded in L 2 (D), it converges weakly in L 2 (D) to ∇T k u for k > 0 and v = ∇u.Thus, u n → u, ∇u n → ∇u in measure.Since we already know that ∇u n → ∇u in measure, it follows from this that ∇u n → ∇u in L q (D) for q < d/(d − 1), and hence, by Poincaré's inequality, that u n → u in W 1,q 0 (D), q ∈ [1, d/(d − 1)).From (4.12) it follows also that the sequence {f un } is equiintegrable.Hence {f un − f u } is equiintegrable, and consequently, since we know that f un → f u a.e.. Finally, to show that u is an entropy solution to (3.2) let us consider an entropy solution w to the problem Aw = −f u − µ, w| ∂D = 0.
By [13, Corollary 3.2] one can find g, from which in much the same way as in the proof of [1,Lemma 4.1] it follows that {u n } is bounded in the space M p/(2−p) (D).Thus, {u n } is bounded in M q (D) for any q ≥ 1.Moreover, from (4.11) and the proof of [1,Lemma 4.2] it follows that {∇u n } is bounded in M q (D) for q ∈ [1, 2).In particular, it follows that {u n } is bounded in W 1,q 0 (D) for q ∈ [1, d/(d − 1)).From this and (4.13) we get (4.14).The rest of the proof runs as before.
In part (ii) of the following main theorem we use some ideas from [4], where L 1 solutions of non-reflected BSDEs with deterministic terminal time and coefficients satisfying the monotonicity condition are considered.L 1 solutions of similar reflected BSDEs are considered in [19].(i) There exists a quasi-continuous q.e.finite entropy solution u of OP(f, µ, ψ).Moreover, if d ≥ 3 then , for any p ≥ 1, q ∈ [1, 2).In particular, in both cases, u ∈ W 1,q 0 (D) for any q ∈ [1, d/(d − 1)).
D) be a solution to EVI(T n f, µ n , ψ) and let γ n be the obstacle reaction associated with u n , i.e.
By Proposition 4.4, u n is a solution of OP(T n f, µ n , ψ).Furthermore, by (4.18), the fact that (T n f ) un u n ≤ 0 and Proposition 4.3 we have Let us define ξ i as in the proof of Theorem 4.5.Then from which as in the proof of (4.12) it follows that Letting k ↓ 0 in (4.20) we get is the obstacle reaction associated with v. From this and (4.14) it follows in particular that v = u.Therefore u is an entropy solution of the problem (3.5) and By the last statement and (4.19), ∇T k u 2 ≤ Ck, k > 0, so according to the remark following the definition of a solution of the obstacle problem, u is q.e.finite.Before we prove that u satisfies the minimality condition (ii) in the definition of the obstacle problem we will show that for q.e.x ∈ D the triple (Y, Z, K) defined by (4.16) is a solution of RBSDE x (f, µ, ψ) satisfying (4.17).By Proposition 4.1, for each x ∈ D the triple (Y n , Z n , K n ) defined by (4.6) is a solution of RBSDE x (T n f, µ n , ψ), i.e.
Z n s dB s , t ≥ 0, P x -a.s., where R n ∼ µ n .Write c n = (T n f ) un + g n and let C n ∼ c n (x) dx, A ∼ G. Then the above equation takes the form To simplify notation set δΦ = Φ n − Φ n+k , Φ := C, K, Y, Z. On the other hand, since u n → u in W 1,q 0 (D) for q < d/(d − 1) and lim T →∞ u(X D T ) = 0, lim T →∞ u n (X D T ) = 0, we have lim Hence, for every x ∈ D, Integrating the above inequality with respect to ν and using Hölder's inequality we get     for T > 0. We know that Y n t = u n (X D t ), t ≥ 0. Since T k u n → T k u in H 1 0 (D) for k > 0, it follows from [8, Lemma 5.1.2]that there is a subsequence, still denoted by n, such that for every k ∈ N and T > 0,  for every w ∈ H 1 0 (D) ∩ L ∞ (D) with the property that there exist k > 0 and w + , w − ∈ W 1,p 0 (D) ∩ L ∞ (D) with p > d such that w = w + a.e. on the set {u > k} and w = w − a.e. on the set {u > k}.For equivalent definitions of renormalized solutions see [7].The first statement, i.e. that v ≥ u q.e.follows immediately from the fact that the renormalized solution of (2.16) is the entropy solution (see [7,Remark 2.17]).To show (4.37) let us define u n , µ n , γ n as in the proof of Theorem 4.6.Since u n is a weak solution of (4.18), it is a renormalized solution of (4.18).Hence  We know that T k u n → T k u in H 1 0 (D), u n → u in W 1,q 0 (D) for q ∈ [1, d/(d − 1)) and µ n + γ n → µ + γ in M b (D).Therefore letting n → ∞ in (4.38) and using [13, Corollary 3.2] and Lemma 2.5 we get (4.37).
and µ M b (D) = |µ|(D).By M 2 b (D) we denote the space of all measures µ in M b (D) such that µ(B) = 0 for every set B ⊂ D such that cap(B) = 0.By M 2,+ b (D) we denote the subset of M 2 b (D) consisting of all positive measures.It is known that if

which proves the theorem. Corollary 3 . 2 .Remark 3 . 3 .
Assume that f : D × R → R, ψ : D → R are measurable functions and µ ∈ M 2 b (D).If f (z, •) is nonincreasing for a.e.z ∈ D and the process ψ(X) is continuous under P x for some x ∈ D then the solution of RBSDE x (f, µ, ψ) is unique.Assume that f, f , µ, µ satisfy the assumptions of Theorem 3.1.From its proof (with

Proposition 4 . 2 .
7) and repeat step by step arguments following (4.3) in Step 1.Let f, µ satisfy the assumptions of Proposition 4.1 and let u ∈ H 1 0 (D) be a weak solution of the problem Au = −µ.

Proof.
Our method of proof will be adaptation of the proof of [13, Proposition 3.8].Let γ be the obstacle reaction associated with u.By Proposition 4.3, γ ∈ M + b (D) and hence γ ∈ M 2,+ b (D) since γ ∈ H −1 (D).Therefore u is an entropy solution of the problem (3.5).Let γ ∈ M 2,+ b (D) and let v be an entropy solution of the problem Av = −µ − γ, v| ∂D = 0 such that v ≥ ψ q.e. in D. What is left is to show that v ≥ u q.e. in D. By [13, Remark 4.5] there is a sequence {γ 10) with v = ψ * and the fact that Aψ