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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.

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  • $\begingroup$ 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. $\endgroup$ Commented Jun 22 at 2:51
  • $\begingroup$ @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? $\endgroup$
    – UDAC
    Commented Jun 22 at 2:54

2 Answers 2

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

Otherwise, $X_+=\bigcup_n\{n-1<f\leq n\}$ 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 $\{f=\infty\}$.

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  • $\begingroup$ It seems that there is a missing part: $\{f=\infty\}$? $\endgroup$
    – user284331
    Commented Jun 22 at 2:32
  • $\begingroup$ Good catch! I think the argument still goes through with the obvious modification. $\endgroup$ Commented Jun 22 at 2:35
  • $\begingroup$ 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? $\endgroup$
    – UDAC
    Commented Jun 22 at 2:40
  • $\begingroup$ 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 $\{f>t\}$. $\endgroup$ Commented Jun 22 at 2:44
  • $\begingroup$ f satisfying the same conditions. $\endgroup$
    – UDAC
    Commented Jun 22 at 2:45
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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 $\int_{0}^{\infty}\mu(A_{t})dt\geq\int_{0}^{c}\mu(A_{t})dt=\int_{0}^{c}\infty dt=\infty$.

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