# Conditions for using Fubini’s Theorem in this problem

I was looking at the answer of this question:

Show: $\int f\, d\mu=\int\limits_0^{\infty}\mu(\left\{x\in X: f(x)>t\right\})\, dt$

You do not need to use the approximation of $$f$$. You can compute directly. Here is the detail: by Fubini's Theorem, $$\begin{eqnarray*} \int_0^\infty\mu(\{x\in X: f(x)>t\})dt&=&\int_0^\infty\int_{X}\chi_{\{x\in X: f(x)>t\}}(y)\;d\mu(y)\;dt\\ &=&\int_{X}\int_0^\infty\chi_{\{x\in X: f(x)>t\}}(y)\;dt\;d\mu(y)\\ &=&\int_X\int_0^{f(y)}dt\;d\mu(y)\\ &=&\int_Xf(y)\;d\mu(y). \end{eqnarray*}$$

Now, exactly why we can use Fubini Theorem? Because the integral over the product measure is finite? In that case, what is the reason? Because I think we cannot use Tonelli since we dont know if the spaces are $$\sigma$$ finite.

I would appreciate some explanation on why we can use Fubini’s Theorem. Thanks.

• You need to state explicitly that $f$ is non-negative. For example, if $f$ is negative, the LHS is negative but the RHS is always non-negative, which is obviously wrong. Commented Jun 22 at 2:51
• @DannyPak-KeungChan Thanks for the answer. I have another doubt : Suppose that f is integrable on X and the space complete. Would that mean the possibility of applying Fubini without Tonelli?
– UDAC
Commented Jun 22 at 2:54

If there exist $$s with $$\mu(\{s or $$\mu(\{f=\infty\})>0$$, then both integrals in your equality are infinite, and you don't need to prove anything.

Otherwise, $$X_+=\bigcup_n\{n-1 is $$\sigma$$-finite and you can use Tonelli on $$X_+$$. On the set $$X\setminus (X_+\cup\{f=\infty\})=\{f=0\}$$ both integrals are zero, and they are also equal on the nullset $$\{f=\infty\}$$.

• It seems that there is a missing part: $\{f=\infty\}$? Commented Jun 22 at 2:32
• Good catch! I think the argument still goes through with the obvious modification. Commented Jun 22 at 2:35
• Thanks for the answer. I have another question : Now suppose that we knew that $f$ is integrable on $X$ and that the space is complete. How would you use Tonelli/Fubini?
– UDAC
Commented Jun 22 at 2:40
• The same way. Or you are talking about $f$ taking non-positive values? In that case the statement doesn't make a lot of sense, as it depends on the set $\{f>t\}$. Commented Jun 22 at 2:44
• f satisfying the same conditions.
– UDAC
Commented Jun 22 at 2:45

Let $$(X,\mathcal{F},\mu)$$ be a measure space. Let $$f:X\rightarrow[0,\infty]$$ be a measurable function. For each $$c\geq0$$, let $$A_{c}=\{x\in X\mid f(x)>c\}$$. We consider two cases.

Case I: $$\mu(A_{c})<\infty$$ for each $$c>0$$. In this case, $$A_{0}$$ is $$\sigma$$-finite with respect to $$\mu$$ because $$A_{0}=\cup_{n}A_{\frac{1}{n}}$$. Define $$\tilde{\mu}(B)=\mu(A_{0}\cap B)$$, for $$B\in\mathcal{F}$$, then $$\tilde{\mu}$$ is a $$\sigma$$-finite measure on $$(X,\mathcal{F})$$. Moreover, $$\int f\,d\mu=\int f\,d\tilde{\mu}$$ and $$\mu\left(\{x\mid f(x)>t\}\right)=\tilde{\mu}\left(\{x\mid f(x)>t\}\right)$$ for each $$t\in(0,\infty)$$. Applying Fubini Theorem on $$\tilde{\mu}\times\lambda$$, where $$\lambda$$ is the Lebesgue measure restricted on $$(0,\infty)$$, the result follows.

Case II. There exists $$c>0$$ such that $$\mu(A_{c})=\infty$$. Not that $$\int f\,d\mu\geq\int_{A_{c}}f\,d\mu\geq c\mu(A_{c})=\infty$$. Also, for each $$t\in(0,c)$$, we have that $$A_{c}\subseteq A_{t}$$, so $$\mu(A_{t})=\infty$$. It follows that $$\int_{0}^{\infty}\mu(A_{t})dt\geq\int_{0}^{c}\mu(A_{t})dt=\int_{0}^{c}\infty dt=\infty$$.