# lebesgue integral uniform convergence

Let $f_n, f : [a,b] \to R.$

If $f_n \to f$ uniformly then show that the lebesgue integrals are equal. ie. $\int f = lim \int f_n$

This is clearly true for continuous functions, but how do I handle the case of non-continuous functions?

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Do you know how to do this argument for Riemann integrable functions? It's the same one for Lebesgue integrable functions. Hint: if $f_n \rightarrow f$ uniformly, then for all $\epsilon > 0$ and all sufficiently large $n$, $|f_n(x) - f(x)|$ is measurable and less than $\epsilon$ for all $x \in [a,b]$. What does that tell you about the Lebesgue integral $\int_{[a,b]} f_n -f$?

Note that the hypothesis that the total measure of the space be finite is essential. It is easy to construct a sequence of integrable continuous functions $f_n: [0,\infty) \rightarrow [0,\infty)$ which converges uniformly to $0$ but so that $\int_{[0,\infty)} f \rightarrow \infty$.

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does it imply that $\int_{[a,b]} f_n -f$ is at most $(b-a)\epsilon$? – jack Feb 28 '11 at 6:14
Yes, it sure does. (Precisely, it implies that the absolute value of your integral is at most $(b-a)\epsilon$, which is what you want.) – Pete L. Clark Feb 28 '11 at 6:17
@PeteL.Clark What is an easy counterxample to show that the finite measure of the space is necessary? – Nhat Nov 22 '14 at 23:56
@KitKat $f_n$ equal to $1/n$ in $(0, 1)$, to $1/2n$ in $(1, 2)$ and so on. They converge uniformly to $0$ but $\int_{[0, \infty)} f_n = \infty$ for every $n$ – Ant Mar 10 '15 at 15:53

It follows from

$\left| \int f \, dm - \int f_n \, dm \right| \le \int \left|f - f_n\right| \, dm \le (b-a) \Vert f - f_n \Vert_\infty$

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