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Let $f\colon X\to\overline{\mathbb{R}}_{\geq 0}$ be a measurable function. Show that $$ \int f\, d\mu=\int\limits_0^{\infty}\mu(\left\{x\in X: f(x)>t\right\})\, dt. $$ (The right integral is to be read as an Lebesgue-integral.)

Hello, my idea is to approximate $f$ by the functions $$ f_n:=2^{-n}\sum\limits_{k=1}^{\infty}\chi_{\left\{x\in X: f(x)> k/2^n\right\}}. $$

With this approximation by simple functions I got $$ \int\limits_X f\, d\mu=\lim\limits_{n\to\infty}\int\limits_X f_n\, d\mu=\lim\limits_{n\to\infty}\sum\limits_{k=1}^{\infty}\frac{k}{2^n}\cdot\mu\left(\left\{x\in X: f_n(x)=k/2^n\right\}\right)\\=\lim\limits_{n\to\infty}\sum\limits_{k=1}^{\infty}\frac{k}{2^n}\cdot\mu\left(\left\{x\in X:f(x)>k/2^n\right\}\right) $$

This is the point where I do not come along anymore... is that right to this point and if yes: How can I continue in order to get the desired right side?

With regards

2 Answers2

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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*}

xpaul
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  • I thank you for that proof and I will try to understand it. Nevertheless I have to make the proof with the approximation because it's the wish of the professor^^ –  Aug 05 '13 at 13:51
  • Do you have an idea how I can continue my proof with the approximation? –  Aug 05 '13 at 15:01
  • @xpaul to use Fubini, you need to know if $\chi_{{x\in X:f(x)>t}}$ is integrable? But we can use Tonelli's if we assume $X$ to be $\sigma$-finite. – Ruzayqat Dec 21 '14 at 02:19
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If $\phi$ is a nonnegative simple function taking values $0 = c_0 < c_1 < c_2 < \cdots < c_n$ on the disjoint sets $E_0, E_1, \ldots, E_n$, then for $t \ge 0$ you have $$\{\phi > t\} = E_k \cup \cdots \cup E_n$$ whenever $c_{k-1} \le t < c_k$ for all $1 \le k \le n$, and $\{\phi > t\} = \emptyset$ if $t \ge c_n$. Thus $$ \int_0^\infty \mu(\{ \phi > t\}) \, dt = \sum_{k=1}^n \int_{c_{k-1}}^{c_k} \mu(\{ \phi > t\}) \, dt = \sum_{k=1}^n (c_k - c_{k-1}) (\mu(E_k) + \cdots + \mu(E_n)).$$ It is straightforward (i.e. just write it out) that $$ \sum_{k=1}^n (c_k - c_{k-1}) (\mu(E_k) + \cdots + \mu(E_n)) = \sum_{k=1}^n c_k \mu(E_k) = \int_X \phi \, d\mu.$$

So, the result is true for simple functions. Try to carry out the general case using the usual limiting tools.

Umberto P.
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  • The next step is to consider an non-negative measurable function f and to show it for this function. Here I can use that there exists an approximation of f by simple functions $f_n$ and that $\int_f=\lim_{n\to\infty}\int f_n$. Right? –  Aug 05 '13 at 18:12
  • I would simply say: $\int_X f, d\mu=\lim\limits_{n\to\infty}\int_Xf_n, d\mu= \lim\limits_{n\to\infty}\int_0^{\infty}\mu(\left{x\in X:f_n(x)>t\right}, dt$? This is valid because the assumption is shown for simple functions $f_n$. And now what is that limit? Is that the desired right side, because of $\lim\limits_{n\to\infty}f_n=f$? –  Aug 05 '13 at 18:20
  • But WHY is $\lim\limits_{n\to\infty}\int\limits_0^{\infty}\mu(\left{x\in X:f_n(x)>t\right}), dt=\int\limits_0^{\infty}\mu(\left{x\in X:f(x)>t\right}), dt$ - what is the reason? –  Aug 05 '13 at 18:31
  • You could take a sequence ${f_n}$ of simple functions that increases to $f$. Show that $\mu({f_n > t})$ increases to $\mu({f > t})$ and use the monotone convergence theorem. – Umberto P. Aug 05 '13 at 18:36
  • To be honest I do not know exactly how to show $\lim\limits_{n\to\infty}\mu(\left{f_n>t\right})=\mu(\left{f>t\right})$. Using your notation for $f_n$ I only have $\lim\limits_{n\to\infty}\mu (\left{f_n>t\right})=\lim\limits_{n\to\infty}\mu(\bigcup\limits_{j=k}^{n}E_j)$. –  Aug 05 '13 at 18:45
  • Ah, its the (increasing) continity of the measure –  Aug 05 '13 at 19:01