Is my proof true? (problem about operator theory and measure theory) The following is the problem which I wrote a solution. Please look at a glance and leave your comments. Please let me know if there is a flaw in my solution or please make a hint for another solution. 

Problem:
  Let 
  $\mathbb{A}$
  be a measurable subset of 
  $\mathbb{R}$. 
  For every 
  $f \in L^{1}(\mathbb{R})$ 
  and for every 
  $ y \in \mathbb{R}.$ 
  Let
  $$T(f,y)=\int_{\mathbb{A}}f(x-y)dx $$
  (a) Prove that for every 
  $y\in \mathbb{R},$
  the function
  $f\mapsto T(f,y)$
  is continuous on
  $ L^{1}(\mathbb{R}). $
(b) Prove that for every 
  $ f\in L^{1}(\mathbb{R}) $ 
  the function 
  $y\mapsto T(f,y)$
  is continuous on
  $\mathbb{R}.$

Here is what I have done:

(a) Let 
  $f_{n}\to f  $
  on 
  $ L^{1}(\mathbb{R}). $
  Then
  $$\lim\limits_{n\to\infty}\int_{\mathbb{R}}\vert f_{n}- f \vert dx=0 $$
  Now, 
  $$\int_{\mathbb{R}}\vert T(f_{n},y)-T(f,y) \vert dy= \int_{\mathbb{R}}\vert \int _{\mathbb{A}}( f_{n}(x-y)-f(x-y))dx\vert dy $$
  $$\leq \int_{\mathbb{A}}( \int _{\mathbb{R}}\vert f_{n}(x-y)-f(x-y)\vert dy) dx\to 0 $$
  (b)Let 
  $y_{n}\to y,$ we have that
  $$\lim\limits_{n\to\infty}\vert T(f,y_{n})-T(f,y) \vert= \lim\limits_{n\to\infty}\vert \int_{\mathbb{A}}(f(x-y_{n})-f(x-y)) dx\vert$$
  Since 
  $f\in L^{1}(\mathbb{R})$
  so there exist
  $s_{m}\in C_{c}(\mathbb{R})$ 
  such that
  $s_{m}\to f.$
  Therefore, 
  $$ \lim\limits_{n\to\infty} \int_{\mathbb{A}}(f(x-y_{n})-f(x-y))dx=\lim\limits_{n\to\infty}\int_{\mathbb{A}}\lim\limits_{m\to\infty}(s_{m}(x-y_{n})-s_{m}(x-y))dx $$
  $$=\lim\limits_{m\to\infty}\int_{\mathbb{A}}\lim\limits_{n\to\infty}(s_{m}(x-y_{n})-s_{m}(x-y))dx=0 $$

 A: Your proof for 1 is incorrect: You do not know if $T(f,y)$ is in $L^1(\Bbb R)$ so $\int_{Bbb R} |T(f,y)-T(f_n,y)|\,dy$ does not make sense. For example if $f=\chi_{[0,1]}$ and $\Bbb A= \Bbb R$ you have $T(f,y)=1$ which is certainly not integrable.
To correct it:
\begin{align}
|T(f,y)-T(f_n,y)|&=\left|\int_{\Bbb A}f(x-y)-f_n(x-y)\,dx\right|≤\int_{\Bbb A}|f(x-y)-f_n(x-y)|\,dx\\
&≤\int_{\Bbb R}|f(x-y)-f_n(x-y)|\,dx=\|f-f_n\|_{L^1}
\end{align}
This converges to zero so $T$ is continuous in $f$ for fixed $y$.
Your proof for 2 is also incorrect: It is unclear exactly what you want to do but it looks like you want to pull the limit in $n$ past the integral and swap it with the limit in $m$ in the following expression:
$$\lim_n \int_{\Bbb A} \lim_m f(x-y)- s_m(x-y_n)\,dx$$
You cannot do this without comment. I'm not sure about a correct proof at the moment.
A: A possible proof for 2.
$\forall\varepsilon>0$


*

*$\exists\ \delta>0\ s.t. \forall F\subset\mathbb{R} \ \&\ m(F)<\delta$, $\int_F|f(x-y_n)-f(x-y)|dx<\varepsilon/4$ for all $n$.

*$\exists N$ enough large such that $\int_{\mathbb{R}\backslash[-N, N]}|f(x-y_n)-f(x-y)|dx<\varepsilon/4$ for all $n$.

*$\exists$ close set $E\subset\mathbb{R}$ such that $m(\mathbb{R}-E)<\delta$ and $f$ is continuous on $E$.


We can easily get these conclusions by a simple inequality $\int_S|f(x-y_n)-f(x-y)|dx\leq\int_S|f(x-y_n)|dx+\int_S|f(x-y)|dx$, where $S$ is any measurable set. 
let $x_n=y_n-y\to0$, and we have
\begin{align}
&|\int_Af(x-y_n)-f(x-y)dx|\\
& \leq\int_\mathbb{R}|f(x-y_n)-f(x-y)|dx\\
& \leq\varepsilon/4+\int_{[-N, N]}|f(x-y_n)-f(x-y)|dx\\
& =\varepsilon/4+\int_{[-N-y, N-y]}|f(x-x_n)-f(x)|dx
\end{align}
Let $G\triangleq E\cap(E+x_n)\cap[-N-y, N-y]$
So $ [-N-y, N-y]\subset G+[\mathbb{R}-E]+[\mathbb{R}-(E+x_n)]$
Due to conclusion 3
original formula $\leq3\varepsilon/4+\int_G|f(x-x_n)-f(x)|dx$
f is uniform continuous on G. So when $|x_n|$ is small enough
original formula $\leq\varepsilon$
