Let $\mu$ be a finite measure on $(\mathbb R, \mathcal B(\mathbb R)).$ Define $$f:\mathbb R \to \mathbb R, \quad f(x) = \int_\mathbb R e^{-(x-y)^2}\mu(dy).$$ Show that $f(x) \to 0$ as $x \to \infty$.
Proof:
We want to apply dominated convergence. Clearly $\lim_{x\to \infty } e^{-(x-y)^2} = 0$ for each $y\in \mathbb R$. Also, since $(x-y)^2 \geq 0$, we have $\left \lvert e^{-(x-y)^2}\right \rvert \leq 1$ and $1$ is integrable since $\mu$ is finite.
Questions:
1) Is this proof correct?
2) If $\mu$ was the Lebesgue measure, $f(x) = \sqrt{\pi} \neq 0$. Of course this is not a contradiction since $\lambda$ is not finite. But what is the problem here? That we don't find a majorizing integrand?
Yes for the first one, and yes for the second one: the main obstruction are the bumps of your family of functions, which would force the uniform bound to be greater than $1$ everywhere, and so not integrable for the Lebesgue measure.