Let $z(x):=\int_{Y} f(x,y)d\mu(y)$ for $x \in \mathbb R$ be an integral function where $\mu$ is a finite(!) Borel measure on $Y$ and $x \mapsto f(x,y)$ is continuous for every $y.$

Moreover, we know that $\int_{ Y} \left\lvert f(x,y) \right\rvert d\mu(y)$ is uniformly bounded in $x.$

My question is whether this implies that $z$ is continuous as well?

What is the obstacle: It seems that the dominated convergence theorem does not apply so easily as $f(x,y)$ is not known to be uniformly bounded in both $x$ and $y.$

  • 1
    $\begingroup$ This is, in disguise, the fact that pointwise convergence plus $L^1$ bounded does not imply $L^1$ convergence. $\endgroup$ – Nate Eldredge Apr 22 '18 at 17:09

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The answer is no. E.g., let $\mu(dy)=g(y)dy$ and $$f(x,y)=\frac1{x^2}g\Big(\frac{x-y}{x^2}\Big)$$ if $x\ne0$, and $f(0,y)=0$ for all real $x$ and $y$, where $g$ is the standard normal density: $g(y):=\frac1{\sqrt{2\pi}}\,e^{-y^2/2}$. Then $\mu$ is a probability measure on $Y=\R$, $x \mapsto f(x,y)$ is continuous for every $y$, and \begin{equation} z(x)=\int_\R |f(x,y)| \mu(dy) \end{equation} is $1$ for $x\ne0$ and $0$ for $x=0$.


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