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I want to find some sequence $\{f_n\}$ of Lebesgue integrable functions on an interval $I$ such that the pointwise limit exists almost everywhere on $I$, the limit function $f$ is Lebesgue-integrable on $I$ but: $$\int_I f \neq \lim_{n\to\infty}\int_I f_n.$$ In Dominated Convergence Theorem we have that if $f_n$ is dominated by some Lebesgue integrable function $g$ we don't even need the hypothesis of $f$ being Lebesgue-Integrable. But what happens if we add it but remove the hypothesis of $g$ dominating the sequence?

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  • $\begingroup$ Let $f_n=n1_{[0,1/n]}$, then $f_n$ conveges to $f=0$, but $\int f_nd\mu=1$. $\endgroup$ – Pei-Lun Tseng Apr 18 '17 at 0:51
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You can always look at the classical $f_n(x) = nx^n$ on $(0,1).$

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Let the underlying space be $(0,1)$ with Lebesgue measure. Let $f_n=n\cdot 1_{(0,1/n)}$. Then $f_n\to 0$ pointwise, but $\int_0^1 f_n=1$.

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  • $\begingroup$ You mean pointwise a.e. $\endgroup$ – zhw. Apr 18 '17 at 1:07
  • $\begingroup$ @zhw. Corrected, thanks. $\endgroup$ – shrinklemma Apr 18 '17 at 1:09
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For an example consisting of continuous functions on a closed interval and pointwise convergence everywhere (to a continuous function), consider $f_n:[0,1]\to \mathbb{R}$ given by a graph of a triangle with one foot on $0$, basis $\frac{1}{2^n}$ and height $2^n$, and $0$ everywhere else.

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