Lebesgue integral question from wiki

Question

I have started studying Lebesgue integration and I have a question regarding the Lebesgue integral.

In the Wikipedia entry on "Lebesgue integration" they define the Lebesgue integral as:

Let $f: \mathbb{R} \rightarrow \mathbb{R}^{+}$ be a positive real-valued function. $$ \int f d\mu = \int_{0}^{\infty}f^{*}(t)dt $$ where $f^{*}(t) = \mu(\{x |f(x) > t\})$.

The Lebesgue integration notes that I am studying define the Lebesgue integral of a positive measurable function as $$ \int f d \mu = \text{sup}\left\{ \int \phi d\mu :\; \phi \text{ is a simple function and } 0 \leq \phi \leq f \right\} $$ I want to know if this wiki definition is equivalent to the integral constructed from simple functions: if so, how can this be easily shown?

Answer 1

Here is a proof based on the so-called "standard machinery" of measure theory : we first prove the result for $f$ simple, and we extend to general $f$ by limit theorems.

Let $f:\mathbb R\to\mathbb R_{\ge0}$ be measurable, we want to show that $$\text{sup}\left\{ \int \phi d\mu :\phi \text{ is simple and } 0 \leq \phi \leq f \right\} = \int_{0}^{\infty}f^{*}(t)dt\tag1$$ where $f^{*}:t\mapsto \mu(\{x \mid f(x) > t\}) $.

First, note that since $f^*(t)\ge 0$ for all $t$, we have that for any sequence $(T_n)_n$ monotonically increasing to $\infty$, the sequence $\Big(\int_0^{T_n}f^*(t) dt \Big)_n$ is non-decreasing and so the integral $\int_{0}^{\infty}f^{*}(t)dt $ is well defined as an element of $\mathbb R\cup \{\infty\}$.

$\bullet$ We will first show the result for $f$ simple : let $c_1,\ldots,c_n\ge 0$ and $A_1,\ldots,A_n\subset\mathbb R$ be disjoint measurable sets such that $f=\sum_{i=1}^n c_i\mathbf 1_{A_i}$. Since there are finitely many of them, we can also assume w.l.o.g. that $0=:c_0\le c_1\le\cdots\le c_n$. In that case we have by definition that $$\text{sup}\left\{ \int \phi d\mu :\phi \text{ is simple and } 0 \leq \phi \leq f \right\} = \sum_{i=1}^n c_i\mu(A_i), $$ and so the equality we want to show reduces to $$\sum_{i=1}^n c_i\mu(A_i) = \int_0^\infty f^*(t) dt.\tag2 $$ But now note that for any given $t\ge0$, \begin{align}f^*(t):=\mu(\{x \mid f(x) > t\}) &= \mu\left(\bigcup_{i \text{ such that }c_i>t} A_i\right)\\ &=\sum_{i \text{ such that }c_i>t}\mu(A_i).\end{align}

And so we see that for any $1\le k\le n$, we have $$\int_{c_{k-1}}^{c_k}f^*(t)dt=\int_{c_{k-1}}^{c_k}\left(\sum_{i=k}^n\mu(A_i)\right)dt = (c_k-c_{k-1})\sum_{i=k}^n\mu(A_i), $$ Which finally gives $$\int_0^\infty f^*(t) dt = \int_{c_0}^{c_n} f^*(t) dt = \sum_{k=1}^n (c_k-c_{k-1})\sum_{i=k}^n\mu(A_i)= \sum_{i=1}^n c_i\mu(A_i),$$ (the last equality can be shown e.g. by induction on $n$) which is precisely the equality $(2)$ we wanted to show.

$\bullet$ Now if $f$ is not assumed to be simple anymore, we know that there exists a sequence $(f_n)_n$ of simple functions such that $f_{n}\le f_{n+1}$ for all $n$ and $f_n(x)\to f(x)$ for all $x\in\mathbb R$. Furthermore, if we denote by $f^*_n$ the function $t\mapsto\mu(\{x \mid f_n(x) > t\})$, we also have that $f^*_{n}\le f^*_{n+1}$ for all $n$ and $f^*_n(x)\to f^*(x)$ for all $x\in\mathbb R$.

It now follows from the monotone convergence theorem applied to both of these sequences (technically, we respectively apply the Lebesgue integral and Riemann integral version of the MCT) that $$\int fd\mu = \lim_{n\to\infty}\int f_n d\mu =\lim_{n\to\infty}\int_0^\infty f^*_n(t)dt = \int_0^\infty f^*(t)dt, $$ which completes the proof.

Answer 2

A hint: break the Riemann integral into the limit of a summation, and it should be clear that they are the same. $f^*(t)dt$ is the area on the strip between two step functions that are a distence $dt$ apart.

Answer 3

Let $X$ be a set and $\mathcal S$ be a $\sigma$-algebra on $X$. Let $\mu$ be a measure on $(X,\mathcal S)$, that is a map from $\mathcal S$ to $[0,\infty]$ such that $\mu(\varnothing)=0$ and $\mu\left(\bigcup_{n\in\omega} A_n\right)=\sum_{n\in\omega}\mu(A_n)$ for any countable family $\{A_n:n\in\omega\}$ of pairwise disjoints members of $\mathcal S$.

Proposition. Let $f:X\to [0,\infty)$ and $f^*:(0,\infty)\to [0,\infty)$ be functions such that $f^{*}(t) = \mu(\{x |f(x) > t\})$ for each $t>0$. Let $I=\int_{0}^{\infty}f^{*}(t)dt$ and $S$ be the supremum of all sums of the form $\sum_{i=1}^n t_i\mu(A_i)$, where $A_1,\dots, A_n\in \mathcal S$ are pairwise disjoint and for each $i\le n$ we have $\mu(A_i)<\infty$ and $f(x)\ge t_i>0$ for each $x\in A_i$. Then $I=S$.

Proof. The function $f^{*}$ is nonnegative and nondecreasing, so the improper Riemann integral $I=\int_{0}^{\infty}f^{*}(t)dt$ exists (but can be infinite).

Suppose first that $S<\infty$. Let $\varepsilon>0$ be any number. Then there exist natural $n$, pairwise disjoint sets $A_1,\dots, A_n\in \mathcal S$, and positive numbers $t_1<\cdots<t_n$ such that for each natural $i\le n$ we have $\mu(A_i)<\infty$ and $f(x)\ge t_i>0$ for each $x\in A_i$, and $\sum_{i=1}^n t_i\mu(A_i)>S-\varepsilon$. For the convenience put $t_0=0$ and $t_{n+1}=\infty$. For each natural $i$ put $B_i=\bigcup_{j=i}^n A_j$. Then for each natural $i$ and each $t<t_i$ we have $f(B_i)\subset (t,\infty)$, so $f^*(t)\ge \mu(B_i)$. Then we have $$S-\varepsilon<\sum_{i=1}^n t_i\mu(A_i)=\sum_{i=1}^n (t_i-t_{i-1})\sum_{j=i}^n\mu(A_j)=\sum_{i=1}^n (t_i-t_{i-1})\mu(B_i)=$$ $$\sum_{i=1}^n \int_{t_{i-1}}^{t_i} \mu(B_i)dt\le \sum_{i=1}^n \int_{t_{i-1}}^{t_i} f^*(t) dt = \int_{0}^{t_n} f^*(t) dt\le I.$$ It follows $S\le I$. For the case $S=\infty$ we can show that $S\le I$ verbatim to the above, but the only changes of $S-\varepsilon$ to any $N>0$.

Suppose first that $I<\infty$. Let $\varepsilon>0$ be any number. Then there exist positive numbers $T<T'$ such that $\int_{T}^{T'}f^*(t) dt>I-\varepsilon$. By the definition of the definite integral, there exist a natural number $n$ and positive numbers $T=t_0<\cdots<t_n=T'$ such that $$\int_{T}^{T'} f^*(t) dt-\varepsilon<\sum_{i=1}^n (t_i-t_{i-1})f^*(t_i).$$ Put $B_0=\varnothing$ and for each natural $i\le n$ put $B_i=f^{-1}(t_i,\infty)$. Then $B_i\in\mathcal S$ and $\mu(B_i)=f^*(t_i)<\infty$. Next, put $A_n=B_n$ and for each natural $i\le n-1$ put $A_i=B_i\setminus B_{i+1}$. Then $A_i\in\mathcal S$, $\mu(A_i)\le\mu(B_i)<\infty$, $f(x)>t_i$ for each $x\in A_i$, and $B_i=\bigcup_{j=i}^n A_j$. Now we have

$$\sum_{i=1}^n (t_i-t_{i-1})f^*(t_i)=\sum_{i=1}^n (t_i-t_{i-1})\mu(B_i)=$$ $$ \sum_{i=1}^n (t_i-t_{i-1})\sum_{j=i}^n\mu(A_j)=\sum_{i=1}^n (t_i-t_0)\mu(A_i)\le S-\varepsilon.$$ It follows $I\le S$. For the case $I=\infty$ we can show that $I\le S$ verbatim to the above, but the only changes of $I-\varepsilon$ to any $N>0$. $\square$

Answer 4

Define $\int f d\mu$ as usual. (I.e. Supremum of $\sum x \mu(s^{-1}(x))$ where $s$ is a simple measurable function such that $s≦f$.)

Note that $f^*$ is almost everywhere continuous since it is monotonically decreasing. Moreover, it is continuous. (Why?)

So $\int_0^\infty f^*(t)dt$ is well-defined where the integral here is Riemann.

Since $f^*$ is continuous, as Stella mentioned, it can be represented as a summation.

Moreover, do you know a theorem :"For every nonnegative measurable function, there exists a monotonically increasing sequence of simple measurable functions converging to that measurable function"?

Construction in that proof shows that the above summation is exactly the Lebegue integral of $f$.