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Consider a function $f(x,y):\mathcal{X}\subseteq \mathbb{R}\times \mathcal{Y}\subseteq \mathbb{R}\rightarrow \mathbb{R}$ and suppose that it is continuous over $y$.

Assume also that $\int_{\mathcal{X}} f(x,y) dx$ exists and is finite.

Can we say that $\int_{\mathcal{X}} f(x,y) dx$ is still continuous over $y$, or I need to impose other restrictions?

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    $\begingroup$ How about $f : [0,\infty) \times [0,\infty) \to \mathbb{R}$ defined by $$ f(x, y) = y \exp(-yx)? $$ It satisfies $$ \int_{[0,\infty)} f(x, y) \, dx = \begin{cases} 1, & y > 0 \\ 0, & y = 0 \end{cases}. $$ $\endgroup$ Aug 29, 2017 at 18:12

2 Answers 2

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The answer is no.

For instance, note that Fourier analysis tells us that there exists a sum of continuous functions $g_k(x)$ over $[0,1]$ such that the sum $g(x) = \sum_{k=1}^\infty g_k(x)$ is discontinuous (the link is to the Fourier series of a square wave). With that in mind, define $$ f: [0,1]\times[0,1]\to \Bbb R\\ f(x,y) = k(k+1)g_k(y) \quad \text{for all } x \in (1/(k+1),1/k] $$ You will find that $\int_0^1 f(x,y)\,dx = g(y)$

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Here is a simple counterexample: Take $f(x,y) = |y|\max(0,1-|xy|)$, then $\int f(x,y) dx = \begin{cases} 1 , & y\neq 0 \\ 0, & y=0 \end{cases}$

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