Continuity of a function (Real Analysis)

Question

In Stochastic process, Bass claim (without proof) the function

$f:\mathbb{R} \to \mathbb{R}$ defined by $f(x)=\int_{A} e^{\tfrac{-(y-x)^2}{2t}}dy$ is continuous, where $A$ is a Borel measurable set.

He says use dominated convergence theorem, it is clear. But I get confused here. Because, HOW?

I tried to use the definition of continuity ($\epsilon-\delta$ argument) to prove it, for instance,

$|\int_{A} e^{\tfrac{-(y-x)^{2}}{2t}}dy-\int_{A} e^{\tfrac{-(y-x_{n})^{2}}{2t}}dy|$ $ \leq |\int_{A \cap [y \leq M]} e^{\tfrac{-(y-x)^{2}}{2t}}dy-\int_{A \cap [y \leq M]} e^{\tfrac{-(y-x_{n})^{2}}{2t}}dy|$

+$|\int_{A \cap [y \geq M]} e^{\tfrac{-(y-x)^{2}}{2t}}dy-\int_{A \cap [y \geq M]} e^{\tfrac{-(y-x_{n})^{2}}{2t}}dy|$

The first term I think we can use the dominated convergent theorem, and the second one use the fact that the integrand is $L^{1}$-integrable, but I can't give a good $\epsilon-\delta$ argument, can someone help?

Furthermore, if the $f$ becomes $f(x)=\int_{A} F(y)e^{\tfrac{-(y-x)^2}{2t}}dy$ where $F$ is bounded measurable, will it be continuous still? It is important since it might gives a sufficient condition for Strong Markov Process but the author never mention. Thanks!

Answer 1

Instead of $\epsilon-\delta$ approach, try to prove following : For any sequence $x_n$ that converges to $x$, $f(x_n)$ converges to $f(x)$.

Now, take an arbitrary such sequence. Then, define $g_n(y)= e^{\frac{-(y-x_n)^2}{2t}}$ and $g(y)=e^{\frac{-(y-x)^2}{2t}}$. At last, use DCT to conclude that $f(x_n)$ converges to $f(x)$.