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Let $(f_n)$ be a sequence of functions defined in $[a,\infty)$, which uniformly converges to $f$ in every interval $I_b$ of the form $[a,b]$. Assume every function in the sequence is integrable in $I_b$ for all $b\geq a$. Finally, assume $\int_a^\infty f(x)dx$ and $\int_a^\infty f_n(x)dx$ converge for all n. Does the following equality necessarily hold? $$\lim_{n\rightarrow \infty}\int_a^\infty f_n(x)dx = \int_a^\infty f(x)dx$$ First, since the sequence converges unformly, we have $$\lim_{n\rightarrow \infty}\int_a^b f_n(x)dx = \int_a^b f(x)dx$$ Second, $$\lim_{n\rightarrow \infty}\int_a^\infty f_n(x)dx = \lim_{n\rightarrow \infty}\left[\lim_{b\rightarrow \infty}\int_a^b f_n(x)dx\right]$$ and also $$\int_a^\infty f(x)dx = \lim_{b\rightarrow \infty}\left[\lim_{n\rightarrow \infty}\int_a^b f_n(x)dx\right]$$ So it kind of boils down to this change of limits, and this is what I cannot justify, nor disprove for this case. Any directions? Thanks.

Edit: my current intuition is that the statement is false, and to hold there must be added an extra condition such as a dominating function $g$ for $f_n$, as seen elsewhere in similar questions, but I'm still unable to disprove this equality.

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Indeed, without further assumptions, the result is not true. For example if $f_n(x)=1$ for $x\in [n,n+1)$ and $0$ otherwise, we have the uniform convergence on each bounded interval to the constant function equal to $0$, but $\int_1^{+\infty}f_n(x)\mathrm dx=1$ for each $n$.

If we furthermore assume that $g(x):=\sup_{n\geqslant 1}\left|f_n(x)\right|$ is integrable on $[a,+\infty)$, then we get the result.

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    $\begingroup$ Thanks, that was a helpful counterexample. $\endgroup$ – Yoni Jun 8 '16 at 22:39

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