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A follow up by this question, I was wondering what if we replace the uniform point-wise boundness of $\{f_n\}$ by uniform convergence of $f_n\rightarrow f$, can we still obtain the convergence of the integral $\lim_{n\rightarrow\infty}\int_Af_nd\mu\rightarrow\int_Afd\mu$ in some weak form?

Generally, if we do not have any control on the convergence of $f_n$, the integral might not converge because $f_n$ might be very different from $f$ on some set $A'$ such that the integral of $\mid f_n-f\mid$ will not become smaller on a set $A'$ with positive measure. But by uniform convergence, I could circumvent this kind of problem.

quallenjäger
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If I understand your question is :

Let $\{f_n\}$ be a sequence of measurable functions on a set of finite measure $E$ such that $f_n \rightarrow f$ uniformly. Do we have $\lim\limits_{n \to \infty} \int_E f_n = \int_E f$?

The answer is yes. Indeed: $$\left|\int_E (f_n - f)\mathrm d \mu\right| \le \mu(E) \left\| f_n - f \right\|_\infty$$

Kroki
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  • Oh right! How could I not have seen that. Thanks. – quallenjäger Apr 11 '18 at 17:28
  • However this does not work if $\mu (E) = \infty$. Indeed, let $f_n(x) = e^{-x} + \frac1n$ which converges uniformly to $f(x) = e^{-x}$ on $E = (0,+\infty)$. However, $\int_E f_n = +\infty$ and $\int_E f = 1$ – Kroki Apr 11 '18 at 17:46