# An equality concerning the Lebesgue integral

Let $f:X\mapsto[0,+\infty)$ be a non-negative measurable function defined on the space $X$, endowed with the complete $\sigma$-additive, $\sigma$-finite, measure $\mu$ defined on the $\sigma$-algebra of the measurable subsets of $X$.

I have read that the following equality holds for the Lebesgue integral:

$$\int_X f d\mu = \int_{[0,+\infty)} \mu(\{x\in X: f(x)>t\}) d\mu_t$$where $\mu_t$ is the usual Lebesgue linear measure.

I would like to understand why this equality holds, but I have got serious problems in proving even the measurability of the function $\phi:t\mapsto \mu(\{x\in X: f(x)>t\})$ (necessary for the Lebesgue integral to be defined) to myself, which would be proved if we could verify that, for any $c\in\mathbb{R}$, the set $$\{t\in\mathbb{R}_{\ge 0}:\mu(\{x\in X: f(x)>t\})<c\}$$is measurable. How can we prove the equality (including the measurability of $\phi$)? I $\infty$-ly thank anyone answering.

• $\{t\in\mathbb{R}_{\ge 0}:\mu(\{x\in X: f(x)>t\})<c\}$ is an interval. In the $\sigma$-finite case you can prove the equality using Tonelli's theorem. In general you can do it by first showing it holds for simple functions $f$ and then applying the definition of the integral of a positive function. – David C. Ullrich Jul 6 '16 at 15:22
• @DavidC.Ullrich Thank you for the comment. Yes, I intend the $\sigma$-finite case... – Self-teaching worker Jul 6 '16 at 16:39

The assumptions that $\mu$ is complete and $\sigma$-finite are not necessary. The result holds for any measure $\mu$.

The answer by zhw shows that the function $\phi$ is measurable, so I won't duplicate that here.

Let's first prove the desired equality assuming that $f$ is a nonnegative, real-valued simple function. In this case, we have $$f(x) = \sum_{m=1}^{M}a_m \chi_{A_m}(x)$$ where each $a_m$ is in $[0,\infty)$, and the $A_m$'s are pairwise disjoint and $\mu$-measurable. Observe that for any $t \in [0,\infty)$, $$\{x \in X : f(x) > t\} = \bigcup_{\{m\ :\ a_m > t\}}A_m$$ so $$\mu\{x \in X : f(x) > t\} = \sum_{\{m\ :\ a_m > t\}}\mu(A_m) = \sum_{m=1}^{M}\mu(A_m) \chi_{[0,a_m)}(t)$$ Integrating both sides gives us \begin{aligned} \int_0^{\infty} \mu\{x \in X : f(x) > t\}\ dt &= \sum_{m=1}^{M}\mu(A_m) \int_0^{\infty}\chi_{[0,a_m)}(t)\ dt\\ &= \sum_{m=1}^{M}\mu(A_m) a_m \\ &= \int f d\mu \\ \end{aligned} Therefore the result holds when $f$ is a simple function. To complete the proof for the general case, let $(f_n)$ be an increasing sequence of simple functions which converges pointwise to $f$. For fixed $t$, observe that if $m \leq n$, then $$\{x \in X : f_m(x) > t\} \subseteq \{x \in X : f_n(x) > t\}$$ Moreover, $$\{x \in X : f(x) > t\} = \bigcup_{n=1}^{\infty}\{x \in X : f_n(x) > t\}$$ These two facts imply that $$\mu \{x \in X : f(x) > t\} = \lim_{n \to \infty}\mu \{x \in X : f_n(x) > t\}$$ We may now apply the monotone convergence theorem (twice) to conclude that \begin{aligned} \int_X f\ d\mu &= \int_X \lim_{n \to \infty}f_n\ d\mu \\ &= \lim_{n \to \infty} \int_X f_n\ d\mu \\ &= \lim_{n\to \infty} \int_0^{\infty} \mu \{x \in X : f_n(x) > t\}\ dt \\ &= \int_0^{\infty} \lim_{n \to \infty} \mu \{x \in X : f_n(x) > t\}\ dt \\ &= \int_0^{\infty} \mu\{x \in X : f(x) > t\}\ dt \\ \end{aligned} as desired.

• Thank you so much! Not trivial at all for me: very, very interesting! – Self-teaching worker Jul 6 '16 at 22:05

For measurability: Verify that in your setup (assuming $\mu$ is positive), $t\to \mu(\{f>t\})$ is decreasing on $[0,\infty).$ Then recall that any monotone function on $[0,\infty)$ is Lebesgue measurable there.

For intuition: Consider the graph of some nice positive function $f$ defined on an interval $[a,b].$ Look at the area $\int_a^b f(x)\,dx$ geometrically and you'll see, by slicing horizontally, that it's also equal to $\int_c^d m(\{f>y\})\, dy,$ where $[c,d]$ is the range of $f.$ The picture is helpful here.

• Thank you so much: you comment about the measurability of $\phi$ is enlightening. As to the integral equality, how to prove it if $X$ is the generic measure space described in the OP? Thank you a lot again! – Self-teaching worker Jul 6 '16 at 16:38
• @Self-teachingworker Do you know Tonelli's theorem? – David C. Ullrich Jul 6 '16 at 16:43
• I don't think you need Fubini/Tonelli for this (and it's true even if $X$ is not $\sigma$-finite). It's not hard to prove the equality directly if $f$ is a simple function. Then for the general case, let $f_n$ be an increasing sequence of simple functions which converges pointwise to $f$, and apply the monotone convergence theorem. – Bungo Jul 6 '16 at 17:36
• @Self-teachingworker Suppose $0\le a_1, < a_2 < \dots < a_n$ and $A_1, \dots , A_n$ are disjoint and measurable. Let $f = \sum a_k\chi_{A_k}.$ For $t > a_n, \mu(\{f>t\}) = 0.$ For $a_{n-1}<t\le a_n, \mu(\{f>t\}) =\mu(A_n).$ For $a_{n-2}<t\le a_{n-1}, \mu(\{f>t\}) =\mu(A_n) +\mu(A_{n-1}),$ etc. – zhw. Jul 6 '16 at 18:52
• @Self-teachingworker I've added a full answer for completeness. – Bungo Jul 6 '16 at 19:09

Uisng Fubini, we have that \begin{align*} \int_X fd\mu &= \int_X \int_0^{f(x)}dt d\mu\\ &=\int_X \int_0^{\infty}\mathbb{I}_{\{f(x) > t\}}dt d\mu\\ &=\int_0^{\infty} \int_X \mathbb{I}_{\{f(x) > t\}} d\mu dt\\ &=\int_0^{\infty}\mu(x\in X, f(x) > t) dt. \end{align*} Here, for clarity, we use $dt$ to denote the Lebesgue measure.

• Thank you very much! A major problem in understanding your answer is that I don't know how to prove that the function $(t,x)\mapsto \mathbb{I}_{\{f(x)>t\}}$ is measurable and to prove that $\int_0^{f(x)}dt=\int_0^{\infty}\mathbb{I}_{\{f(x) > t\}}dt$ (if that is the case)... – Self-teaching worker Jul 6 '16 at 22:09
• @Self-teachingworker $$\mathbb I_{f(x) > t}(t) = \begin{cases}1 & \text{if }t \in [0, f(x)) \\ 0 & \text{otherwise} \\ \end{cases}$$ so $\int_0^{\infty}\mathbb I_{f(x)>t}(t)\ dt = \int_{0}^{f(x)} 1\ dt + \int_{f(x)}^{\infty}0\ dt = \int_0^{f(x)}\ dt$. – Bungo Jul 6 '16 at 23:13
• @Bungo Thank you so much!!! The measurability of $(t,x)\mapsto\mathbb{I}_{\{f(x)>t\}}$ is all that remains to me to understand. It's enough to see that $\{(t,x)\in\mathbb{R}_{\ge 0}\times X:\mathbb{I}_{\{f(x)>t\}}<c\}$ is measurable for $c\in (0,1]$ (if $c\le 0$ or $c>1$ it's trivial), but it isn't so straightforward to my eyes... – Self-teaching worker Jul 7 '16 at 9:22
• @Self-teachingworker $\mathbb I_{f(x) > t}$ is a characteristic function, so if $0 < c < 1$, then $\mathbb I_{f(x) > t} < c$ if and only if $\mathbb I_{f(x)>t} = 0$, if and only if $f(x) \leq t$, if and only if $f(x) - t \leq 0$. Now observe that $f(x) - t$ is the sum of the functions $(t,x) \mapsto f(x)$ and $(t,x) \mapsto -t$, both of which are measurable. – Bungo Jul 7 '16 at 16:24
• @Self-teachingworker No problem, this site is a perfect place to ask questions/get feedback when you are self-learning. I don't much care for Kolmogorov-Fomin (especially without an instructor) because as you said, it leaves out far too many details and leaves important results as exercises. The edition translated by Silverman also contains many errors. For measure and integration, a really good introduction is Bartle, The Elements of Integration and Lebesgue Measure Unfortunately it's quite expensive. – Bungo Jul 7 '16 at 19:47