# Is the limit of integrals on the sets with Lebesgue measure converging to zero always zero?

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 $$\int_{E}f dm=\infty$$.

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 $$\int_{E}fdm =+\infty$$.

• Ramiro: If in addition we define $f$ at zero by, e.g., $f(0)=0$, this example remains to be true, isn't that? – serenus Jan 4 '20 at 15:23
• And, why the value of the limit in your example is $\infty$? – serenus Jan 4 '20 at 15:31
• @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. – Ramiro Jan 4 '20 at 15:47
• @serenus $\lim_{n\longrightarrow \infty}\int_{E_n}fdm= +\infty$ because for all $n$, $\int_{E_n}fdm= +\infty$. – Ramiro Jan 4 '20 at 15:51
• Actually I cannot see why the integral on each $E_n$ is $\infty$. – serenus Jan 4 '20 at 15:54

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 $$0$$.

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