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My professor posed a question to us last week about the limit of a function as $k \to \infty$. He asked us to prove that $\int \lim_{k \to \infty} f(x)e^{-\frac{x^2}{k}} dx = \int f(x)dx$. This seems fairly basic, since it can be directly shown that the exponential part of the integrand, $e^{-\frac{x^2}{k}} \to 1$ as $k \to \infty$. Is there something I am missing? Are there extra steps needed to show that the integral of the limit is equal to the limit of the integral? Is this even necessary? I feel like this question is too simple relative to the rest of the material in class, but I'm not sure what I am missing.

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well, $f=g$ implies $\int f=\int g$... – leo Nov 20 '12 at 21:29
Are you sure he didn't ask $\lim_{k\rightarrow \infty}\int f(x)e^{-x^2/k}\,dx$? In that case, you need to justify switching the integral and the limit. – asmeurer Dec 3 '12 at 5:12

Over an interval or compact set, $f(x) e^{-x^2/k} \to f(x)$ uniformly. Then you can get $\int \lim = \lim \int$ from Riemann integration.

In a Lebesgue theory class, looks like you're integrating with uniform measure on some measurable set. So perhaps you need dominated convergence.

Indeed, $f(x) e^{-x^2/k} \to f(x)$ pointwise and $|f(x) e^{-x^2/k}| \leq f(x)$. Therefore

$$ \lim_{k \to \infty} \int f(x) e^{-x^2/k} \, dx \to \int f(x) \, dx$$

So you just need to be sure everything is integrable.

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