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Let $f$ be an extended real-valued Lebesgue measurable function on a set $E$ with its Lebesgue measure $m(E)>0$. Suppose that $(E_n)_{n=1}^{\infty}$ is a sequence of Lebesgue measurable subsets of $E$ with $\lim_{n\longrightarrow \infty} m(E_n)=0$. Is there an example of a non-negative function on $E$ such that $\lim_{n\longrightarrow \infty}\int_{E_n}fdm\neq 0$? Of course, for such a function $f$, we should have $\int_{E}f dm=\infty$.

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Yes, there are such examples. Here is one of them: $f: (0,1) \to \mathbb{R}$ defined by $f(x) = \frac{1}{x}$. Let $E=(0,1)$ and, for $n \geqslant 1$ , $E_n=(0,\frac{1}{n})$.

Clearly $(E_n)_{n=1}^{\infty}$ is a sequence of Lebesgue measurable subsets of $E$ with $\lim_{n\longrightarrow \infty} m(E_n)=0$ and $f$ is a non-negative function on $E$ such that $\lim_{n\longrightarrow \infty}\int_{E_n}fdm= +\infty \neq 0$.

Remark: Let us prove that for any natural $n \geqslant 1$, $\int_{E_n}fdm= +\infty$.

Given any fixed $n \geqslant 1$, note that, for all real $\epsilon \in E_n$, we have $(\epsilon, \frac{1}{n}) \subset E_n$. Since $f$ is non-negative, we have:

$$\int_{E_n}fdm \geqslant \int_{(\epsilon, \frac{1}{n})}fdm=\int_\epsilon^ \frac{1}{n}fdm= \ln\left(\frac{1}{n}\right)-\ln(\epsilon)$$

But $\ln\left(\frac{1}{n}\right)-\ln(\epsilon)$ can be made arbitrarily large, by chosing $\epsilon$ sufficiently close to $0$. So we have that $\int_{E_n}fdm =+\infty$.

Since $E_1=E$, whe have also proved that $\int_{E}fdm =+\infty$.

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  • $\begingroup$ Ramiro: If in addition we define $f$ at zero by, e.g., $f(0)=0$, this example remains to be true, isn't that? $\endgroup$ – serenus Jan 4 '20 at 15:23
  • $\begingroup$ And, why the value of the limit in your example is $\infty$? $\endgroup$ – serenus Jan 4 '20 at 15:31
  • $\begingroup$ @serenus Yes, a second slightly different example is: $E=[0,1)$ and, for $n \geqslant 1$ , $E_n=[0,\frac{1}{n})$ and $f: [0,1) \to \mathbb{R}$ defined by $f(x) = \frac{1}{x}$ if $x \neq 0$ and $f(0)=0$. And yet a third example is: $E=[0,1]$ and, for $n \geqslant 1$ , $E_n=[0,\frac{1}{n}]$ and $f: [0,1] \to \mathbb{R}$ defined by $f(x) = \frac{1}{x}$ if $x \neq 0$ and $f(0)=0$. They are just small variations of the first example. $\endgroup$ – Ramiro Jan 4 '20 at 15:47
  • $\begingroup$ @serenus $\lim_{n\longrightarrow \infty}\int_{E_n}fdm= +\infty$ because for all $n$, $\int_{E_n}fdm= +\infty$. $\endgroup$ – Ramiro Jan 4 '20 at 15:51
  • $\begingroup$ Actually I cannot see why the integral on each $E_n$ is $\infty$. $\endgroup$ – serenus Jan 4 '20 at 15:54
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No. Let $f_n=f.\chi_{E_n}$. Then $\int_{E_n}f\,\mathrm dm=\int_Ef_n\,\mathrm dm$. So, by the dominated convergence theorem (you have $(\forall n\in\mathbb N):\lvert f_n\rvert\leqslant\lvert f\rvert$),\begin{align}\lim_{n\to\infty}\int_{E_n}f\,\mathrm dm&=\lim_{n\to\infty}\int_Ef_n\,\mathrm dm\\&=\int_E\lim_{n\to\infty}f_n\,\mathrm dm\\&=0,\end{align}since $f$ is $0$ outside a set with measure $0$.

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  • $\begingroup$ Yes, I did. I will delete my answer then. $\endgroup$ – José Carlos Santos Jan 4 '20 at 14:47

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