limit of the mollifying sequence

Let $$f:\mathbb{R}^2 \rightarrow \mathbb{R}$$ be an $$L^1(\mathbb{R}^2; \mathbb{R})$$ function. Let $$A \subset \mathbb{R}$$ be a measurable set. Let $$\eta: \mathbb{R}\rightarrow \mathbb{R} \in C_c^{\infty}(\mathbb{R}),$$ with support in $$[-1,1]$$ such that $$\int\limits_{\mathbb{R}}\eta(x)dx=1.$$ Then consider the following limit $$\begin{eqnarray} \lim\limits_{\epsilon \rightarrow 0}\frac{1}{\epsilon}\int\limits_{\mathbb{R}} \int\limits_{A} f(x,y)\eta\left(\frac{x-y}{\epsilon}\right)dy dx. \end{eqnarray}$$ What is the value of the limit? How to prove it.

P.S: Define $$\eta_{\epsilon}(x):=\frac{1}{\epsilon}\eta(\frac{x}{\epsilon}).$$ Then $$\eta_{\epsilon} \rightarrow \delta_0.$$ So, I think the limit is, $$\begin{eqnarray} \int\limits_{A} f(x,x)dx. \end{eqnarray}$$ is it correct? How to prove it?

• if you want $\eta_{\epsilon} \rightarrow \delta_0$, then you need to define $\eta_{\epsilon}(x):=\frac1\epsilon\eta(\frac{x}{\epsilon})$. Right now $\eta_{\epsilon} \rightarrow0$ almost everywhere, which means that the limit of the double integral equals $0$ right now. (Also, what is $A$?) – Greg Martin Jul 12 '20 at 7:24
• @Greg Martin..yes you are right.. I have made the necessary changes...Thanks for the comment. Is the result true now? Or do we need some more regularity assumption on the set A – Sameera Jul 12 '20 at 8:07

Change variables $$\frac{x-y}{\epsilon}\rightarrow z$$ then $$\begin{eqnarray} \frac{1}{\epsilon} \int\limits_{\mathbb{R}} \int\limits_{A} f(x,y)\eta\left(\frac{x-y}{\epsilon}\right)dy dx\\=- \int\limits_{\mathbb{R}} \int\limits_{\left(\frac{1}{\epsilon}x-\frac{1}{\epsilon}A\right)\cap [-1,1]} f(x,x-\epsilon z)\eta\left(z\right)dz dx \end{eqnarray}$$ which converges by the dominated convergence theorem to $$\begin{eqnarray} \int\limits_{\mathbb{R}}f(x,x) \left( -\int\limits_{\mathbb{R}} \lim_{\epsilon\rightarrow 0}\mathbb{1}_{\left(\frac{1}{\epsilon}x-\frac{1}{\epsilon}A\right)\cap [-1,1]}(z) \eta\left(z\right)dz\right) dx \end{eqnarray}$$ Since $$\eta$$ is compactly supported in $$[-1,1]$$, the inner integral converges to a number uniformly in $$x$$. If we know that $$[-1,1]\subset A$$ we get a nicer limit: $$\begin{eqnarray} \frac{1}{\epsilon} \int\limits_{\mathbb{R}} \int\limits_{A} f(x,y)\eta\left(\frac{x-y}{\epsilon}\right)dy dx= \frac{1}{\epsilon} \int\limits_{\mathbb{R}} \int\limits_{\mathbb{R}} f(x,y)\eta\left(\frac{x-y}{\epsilon}\right)dy dx \\= \int\limits_{\mathbb{R}} \int\limits_{\mathbb{R}} f(x,x-\epsilon z)\eta\left(z\right)dz dx \end{eqnarray}$$ which converges by the dominated convergence theorem to $$\begin{eqnarray} \int\limits_{\mathbb{R}}f(x,x) dx\,\int\limits_{\mathbb{R}} \eta\left(z\right)dz=\int\limits_{\mathbb{R}}f(x,x) dx. \end{eqnarray}$$