Let $f_n,\,n\geq1$, non-negative functions and $\mathcal{F}$-measurable in $(\Omega,\mathcal{F},\mu)$.

Show that $$\int_\Omega(\sum_{n=1}^{\infty}f_n)d\mu = \sum_{n=1}^{\infty}\int_\Omega f_nd\mu$$

I supposed first that $f_n$ are simple functions, so $f_n(\omega)=\sum_{i=1}^kx_{n,i}I_{A_{n,i}}(\omega),\,\omega\in\Omega$

So, $$\int_\Omega(\sum_{n=1}^{\infty}f_n)d\mu = \sum_{i=1}^k\sum_{n=1}^\infty x_{n,i}\mu(A_{n,i})=\sum_{n=1}^\infty\sum_{i=1}^k x_{n,i}\mu(A_{n,i})=\sum_{n=1}^{\infty}\int_\Omega f_nd\mu$$ But, I can only do that if the sum to infinity converges. My first question is, how do I prove the case when the sum doesn't converge?

Now, for general $f_n$, I can use $$\int_\Omega f_nd\mu = \sup\bigg\{\int_\Omega\psi_n d\mu:0\leq\psi_n\leq f_n, \psi_n: simple\bigg\}$$ Then \begin{split} \int_\Omega\sum_n f_nd\mu &= \sup\bigg\{\int_\Omega\sum_n\psi_n d\mu:0\leq\sum_n\psi_n\leq\sum_nf_n, \psi: simple\bigg\}\\ &=\sum_n\sup\bigg\{\int_\Omega\psi_n d\mu:0\leq\psi_n\leq f_n, \psi_n: simple\bigg\}\\ &=\sum_n\int_\Omega f_nd\mu \end{split} Can I do that? Won't I have the same problem with the non-convergence? Someone could help, please?

  • $\begingroup$ Have you seen the monotone convergence theorem? $\endgroup$ – Quoka Apr 22 '18 at 14:36
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    $\begingroup$ You can change the order of summation since all the terms are positive. You don't need convergence $\endgroup$ – Quoka Apr 22 '18 at 14:42
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    $\begingroup$ Since the $f_n$'s are positive, you will either have convergence or both sides will tend to $+\infty$. It's not an issue $\endgroup$ – Quoka Apr 22 '18 at 14:43
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    $\begingroup$ As pointed out by MathUser_NotPrime, you don't need convergence. Indeed the result follows from Tonelli's theorem $\endgroup$ – Uskebasi Apr 22 '18 at 15:53
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    $\begingroup$ What you have to do is prove the inequalities $\int \sum f_n \leq \sum\int f_n$ and $\int \sum f_n \geq \sum\int f_n$ separately. Your proof will actually end up mimicking the proof of the monotone convergence theorem. I personally like the proof from Folland's real analysis. I think the below document uses the same proof. Hopefully this is helpful :-) maths.tcd.ie/~richardt/MA2224/MA2224-ch3.pdf $\endgroup$ – Quoka Apr 22 '18 at 17:11

Seriously, no MCT? Your proof for nonnegative looks good. For the proof for simple,

For your proof for simple, it comes down to this: Does the switch between

$$\int_\Omega\sum_{n=1}^{\infty} \text{(random variables)} = \sum_{n=1}^{\infty}\int_\Omega \text{(random variables)}$$

build on the switch between $$\sum_{i=1}^k\sum_{n=1}^{\infty} \text{(real numbers)} = \sum_{n=1}^{\infty}\sum_{i=1}^k \text{(real numbers)}$$

The answer is yes, and based on your proof you would seem to know that. Let's expand your proof in more detail. In particular, how did we get your step (a) below?

$$\int_\Omega(\sum_{n=1}^{\infty}f_n)d\mu \stackrel{a}{=} \sum_{i=1}^k\sum_{n=1}^\infty x_{n,i}\mu(A_{n,i})\stackrel{b}{=}\sum_{n=1}^\infty\sum_{i=1}^k x_{n,i}\mu(A_{n,i})\stackrel{c}{=}\sum_{n=1}^{\infty}\int_\Omega f_nd\mu$$


$$ = \int_\Omega\sum_{n=1}^{\infty}\sum_{i=1}^k x_{n,i}1_{A_{n,i}}d\mu$$

$$ = \int_\Omega\sum_{i=1}^k \sum_{n=1}^{\infty} x_{n,i}1_{A_{n,i}}d\mu \tag{a1}$$

$$ = \sum_{i=1}^k \int_\Omega \sum_{n=1}^{\infty} x_{n,i}1_{A_{n,i}}d\mu$$

$$ = \sum_{i=1}^k \sum_{n=1}^{\infty} \int_\Omega x_{n,i}1_{A_{n,i}}d\mu \tag{a2}$$

$$ = \sum_{i=1}^k \sum_{n=1}^{\infty} x_{n,i}\mu(A_{n,i})$$

$$ = \sum_{n=1}^{\infty} \sum_{i=1}^k x_{n,i}\mu(A_{n,i}) \tag{b}$$

$$ = \sum_{n=1}^{\infty}\int_\Omega f_nd\mu \tag{c}$$

If you know $(a1)$, then you know $(b)$. The proof for $(a1)$ is simply that for a given $\omega$, $x_{n,i}1_{A_{n,i}}$'s are simply numbers, either $x_{n,i}$'s or $0$'s. So actually, before we get to asking about $(b)$, how did you do $(a)$, in particular $(a1)$?

Next, we have $(a2)$. You can prove with MCT, but then again, if you could do that, you could've done it from the beginning. Anyway, $(a2)$ should look familiar: it's the original result but with positive multiples for indicator functions instead of simple or nonnegative functions.

Thus, instead of starting from simple and then extending to nonnegative. You should go:

$$\text{indicator} \ \to \ \text{positive multiple of indicator} \ \stackrel{a2}{\to} \ \text{simple} \ \to \ \text{nonnegative}$$

For indicator, we must show that

$$\int_\Omega \sum_{n=1}^{\infty} 1_{A_n}d\mu = \sum_{n=1}^{\infty} \int_\Omega 1_{A_n}d\mu$$

For positive multiple of indicator, we must show that for $x_n > 0$

$$\int_\Omega \sum_{n=1}^{\infty} x_n 1_{A_n}d\mu = \sum_{n=1}^{\infty} \int_\Omega x_n 1_{A_n}d\mu$$

The latter I think is a simple extension of the former. To prove the former, consider $\sum_{n=1}^{\infty} 1_{A_n}$. It is either finite a.s. or infinite a.s.

If it is finite a.s., then there are only finitely many $A_n$'s s.t. $\omega \in A_n$ for almost all $\omega$. Write those $A_n$'s as $A_{n_1}$, ..., $A_{n_k}$. We have for any other $A_n$ that $\mu(A_n) = 0$ and $0 = 1_{A_n}$ a.s. Then

$$\int_\Omega \sum_{n=1}^{\infty} 1_{A_n}d\mu$$

$$ = \int_\Omega \sum_{n=n_1}^{n_k} 1_{A_n}d\mu$$

$$ = \sum_{n=n_1}^{n_k} \int_\Omega 1_{A_n}d\mu$$

$$ = \sum_{n=n_1}^{n_k} \mu(A_n)d\mu$$

$$ = \sum_{n=1}^{\infty} \mu(A_n)d\mu$$

If it is infinite a.s., then $\int_\Omega\sum_{n=1}^{\infty} 1_{A_n}d\mu = \infty$. It remains to show that $\sum_{n=1}^{\infty} \mu(A_n)d\mu = \infty$. If $\sum_{n=1}^{\infty} \mu(A_n)d\mu < \infty$, then by Borel-Cantelli Lemma 1*, $\mu(\limsup A_n) = 0$. Convince yourself that this contradicts $\sum_{n=1}^{\infty} 1_{A_n} = \infty$ a.s.

*Hopefully your proof of BCL1 did not use the non-alternative one here.


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