Exercise 1.4.42 (Counter-examples to Tonelli's theorem)

Question

$
ewcommand{\intx}[1]{\int_{X} #1 \mathop{}\!\mathrm{d}{\mu}}$
Let $(X,\mathcal{B},\mu)$ be a measure space. Give a counterexample to $$\sum_{n=1}^{\infty}\intx{f_n} = \intx{\sum_{n=1}^{\infty}f_n}$$ when $\left(fight)_{n\in\mathbb{N}}$ is absolutely integrable and not unsigned.

Elu Thingol · Accepted Answer

We give an example of convergent absolutely integrable functions whose limit is not absolutely integrable. \par

Consider $([0,+\infty),\mathcal{B},m)$ where $\mathcal{B}$ are the Borel sets on $[0,+\infty)$ and $m$ is the Lebesgue measure. Define $\left(fight)_{n\in\mathbb{N}}$ by
$$ \forall n \in \mathbb{N}: \quad f_n:[0,+\infty)	o\mathbb{C}, x\mapsto n1_{\left[\frac{1}{n},\frac{2}{n}ight]} - 1_{\left[n,n+1ight]}.$$
Then $f_n$ is measurable, and it equals to $1$ whenever $x$ is in the unit inverval, and to $-1$ otherwise. One one hand we have
$$\sum_{n=1}^{\infty} \intx{\left(1_{\left[\frac{1}{n},\frac{2}{n}ight]} - 1_{\left[n,n+1ight]}ight)} = \sum_{n=1}^{\infty} nm\left({\left[\frac{1}{n},\frac{2}{n}ight]}ight) - m\left({\left[n,n+1ight]}ight)  = \sum_{n=1}^{\infty} 0 = 0.$$

On the other hand we have
$$\intx{\sum_{n=1}^{\infty} \left(n1_{\left[\frac{1}{n},\frac{2}{n}ight]} - 1_{\left[n,n+1ight]}ight)} = \intx{\sum_{n=1}^{\infty} n1_{\left[\frac{1}{n},\frac{2}{n}ight]} - \sum_{n=1}^{\infty} 1_{\left[n,n+1ight]}}.$$
The negative part of this function is not integrable (and the positive too):
$$\intx{ \sum_{n=1}^{\infty} 1_{\left[n,n+1ight]}} = \intx{1_{[2,+\infty)}} = \infty  $$
and the integral of the limit, therefore, does not exist.

Exercise 1.4.42 (Counter-examples to Tonelli's theorem)

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Exercise 1.4.42 (Counter-examples to Tonelli's theorem)

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