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Let $f:\Bbb R^n\times \Bbb R^m\to \Bbb R$ be a bounded measurable function, and $\mu$ be a probability measure on $\Bbb R^m$ which is absolutely continuous w.r.t. Lebesgue measure $\lambda$, e.g. $m =\frac{\mathrm d\mu}{\mathrm d\lambda}$ is the density of $\mu$. Define $$ g(x):=\int_{\Bbb R^m} f(x,y)\mu(\mathrm dy) = \int_{\Bbb R^m} f(x,y)m(y)\lambda(\mathrm dy). $$ Under which conditions can we assure that $g$ is a continuous function of $x$? I'm pretty sure this question should have been studied already, as it is pretty basic. I am interested in the case when one can't assume continuity of $f$ as a function of $x$ for fixed $y$, or as a function of $y$ for a fixed $x$.

Clearly, $g$ is not continuous in general. For example, if $f$ does not depend on $y$, then $g(x) = f(x)$ and hence if $f$ is discontinuous so is $g$. The only way to prevent it is to assume that $f$ somehow "always depends" on $y$ in a non-trivial way as it shall help "average" the discontinuities of $f$. At the same time, I am not sure how to formally assume such dependence. Perhaps, claiming that the image measure $(f_x)_*\mu$ is non-atomic for every $x$ shall help.

I hope that assumptions on $\mu$ allow considering such distributions as exponential or uniform (that is, densities that are either $0$ or smooth).

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Denote $h_x(y) = f(x,y)$. Then the dominated convergence theorem tells us that if $\lim_{x\to x_0}h_x = h_{x_0}$ almost surely in $(\Bbb R^m,\mu)$, then $g$ is continuous at $x_0$. Since $f$ is bounded, I believe it suffices that $h_x\to h_{x_0}$ in probability. – Jonathan Y. Jan 22 '14 at 18:33
@JonathanY. thanks, that actually could help. The uniform integrability is a necessary and sufficient assumption for the convergence in probability to imply the convergence of expectations, so we have it for bounded $f$. – Ilya Jan 22 '14 at 21:27

A quite general condition on $f$ is that $f(x,y)$ is continuous as a function of $x$ uniformly in $y$, that is, given $x_0\in\mathbb{R}^m$ and $\epsilon>0$ there exists $\delta$ (depending on $x_0$ and $\epsilon$ but not on $y$) such that $$ |x-x_0|\le\delta \implies |f(x,y)-f(x_0,y)|\le\epsilon\quad\forall y\in\mathbb{R}^m. $$

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Thank you, although I'm interested in the case when $f$ is continuous in neither of two variables, should have specified it from the beginning. Updated OP now. – Ilya Jan 22 '14 at 18:31
The only thing I can think of is if $f(x,y)=\phi(x-y)$. Then $g=\phi\ast\mu$ will be continuous for a large class of functions $\phi$. – Julián Aguirre Jan 22 '14 at 18:36

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