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Let $f : \mathbb{R}^{n} \times \mathbb{R} \to \mathbb{R}_{\geq 0}$ be continuous everywhere in both arguments. I'd like to know that under what condition(s) the following holds $$ \lim_{y \to y_{0}} \int f(x,y) dx = \int f(x,y_{0}) dx $$ for some $y_{0} \in \mathbb{R}$. Is it to do with uniform convergence?

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2 Answers 2

To answer your question about why you need uniform continuity, it might be instructive to look at how such a proof goes. Fix an $\epsilon > 0$, and consider

$\int_A f(x,y) - f(x,y_0)dx$, for some domain $A \subset \mathbb{R}^n \times \mathbb{R}$. We're trying to show that by picking $y$ close to $y_0$ that this integral can be made small. If we only assume that $f$ is continuous, then it's easy to show that for any $\epsilon > 0$ we can pick a $\delta > 0$ so that if $|(x,y) - (x,y_0)| < \delta$ then $|f(x,y) - f(x,y_0)| < \epsilon$. However, the choise of $\delta$ here depends on the particular $x \in \mathbb{R}^n$ we're talking about. In order to push this line of argument through, you need to be able to make some sort of uniform estimate simultaneously on all $(x,y)$ close to $(x,y_0)$, irrelevant of the choice of $x$. This is what uniform continuity means.

This line of reasoning should allow you to develop examples of continuous functions for which the above limit interchange fails. This will probably be a more instructive endeavor than reading other people's intuition.

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Thanks a lot. It's really instructive, mate. –  Navid Noroozi Jan 2 '13 at 5:15

Since $f$ is continuous, I believe that your question is a specific case of point 1 in this question.

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