Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Normally, Lebesgue integral, for positive measures, is defined in the following way. First, one defines the integral for indicator functions, and linearly extend to simple functions. Then, for a general non-negative function $f$ , it is defined as the supremum of the integral of all simple functions less than or equal $f$.

However, I find another definition in the book of Lieb and Loss, "Analysis". Let $f$ be the non-negative measurable function on a measure space $X$, let $\mu$ be the measure, and define for $t >0$, $$S_f(t) = \{x \in X : f(x) >t\},$$ and $$F_f(t) = \mu(S_f(t)) .$$

Note that $F_f(t)$ is now a Riemann integrable function. Now, the Lebesgue integral is defined as:

$$ \int_X f \, d\mu = \int_0^\infty F_f(t) \, dt,$$

where the integral on the right-hand side is the Riemann integral.

Here is the google books link to the definition.

The book gives a heuristic reason why this definition agrees with the usual definition described here in the first paragraph. Now I would like to have a rigorous proof of the equivalence.

share|improve this question
1  
It follows from Fubini's theorem. –  Siminore Aug 23 '12 at 17:00
    
This fact is sketched for probability measure $\mu$ here - the last formula. I wouldn't take it as a definition, though - just as a useful fact. As a definition it seems to be too hard to work with, e.g. linearity in the argument $f$ is not that easy to see. –  Ilya Aug 23 '12 at 17:00
    
This formula has no reason to be true if the underlying measure isn't $\sigma$-finite. –  Ahriman Aug 23 '12 at 17:08
15  
@timur: Indeed. Else, I would be a mere Thomas. –  Doubting Thomas Aug 23 '12 at 18:57
1  
@Ahriman: The formula is true for an arbitrary measure space (when the right hand side is interpreted as a Lebesgue integral and $F_f(t)$ is replaced by the outer measure $F_f(t) = \mu^\ast(S_f(t))$ and holds as stated if $\mu$ is complete). No need to assume $\sigma$-finiteness or anything else on $\mu$. This is proved e.g. in Fremlin's measure theory, volume 2, 252 O. –  t.b. Aug 26 '12 at 9:14
show 4 more comments

2 Answers 2

I'll paraphrase the slicing argument given in Fremlin's Measure theory, Volume 2, 252O, page 220. Since I'm not going to base a theory of integration on this formula, I think it's legitimate to appeal to (very) basic Lebesgue theory (no Fubini, only monotone convergence) to prove the desired identity. I leave it to you to deal with the necessary but easy modifications if you allow extended real-valued functions or partially defined functions:

Let $(X,\Sigma,\mu)$ be a measure space and let $f\colon X \to [0,\infty)$ be a measurable function. Then for the usual Lebesgue integral $\int_X f\,d\mu$ of $f$ over $(X,\Sigma,\mu)$ we have the identity $$ \int_X f\,d\mu = \int_{0}^\infty \mu\{x\in X\,:\,f(x) \gt t\}\,dt $$ where the second integral is a Lebesgue integral of a non-increasing and non-negative function, hence the integrand is continuous up to a countable subset of $[0,\infty)$, and thus its Riemann integral exists in the extended sense and the Riemann and Lebesgue integrals coincide for it.

To prove this, set $E^{n}_k = \left\{x \in X\,:\,f(x) \gt \frac{k}{2^{n}}\right\}$ and put $$ g_n = \frac{1}{2^n} \sum_{k=1}^{4^n} [E_{k}^n] $$ where $[A]$ denotes the characteristic function of $A$. Note that we have $0 \leq f(x) - g_n(x) \lt 2^{-n}$ whenever $0 \leq f(x) \lt 2^n$, so that $g_n(x) \to f(x)$ for every $x \in X$. Also, $g_n(x) \leq g_{n+1}(x)$ for all $x$, so the sequence $(g_n)_{n \in \mathbb{N}}$ increases to $f$ pointwise everywhere. By the monotone convergence theorem we conclude that $$\tag{1} \int_X f\,d\mu = \lim\limits_{n\to\infty} \int_X g_n\,d\mu. $$ Moreover, for every $t \in [0,\infty)$, we have $$ \{x\in X\,:\,f(x) \gt t\} = \bigcup_{n \in \mathbb{N}} \{x \in X\,:\,g_n(x) \gt t\}, $$ so, using the notation in the question, we have for all $t \in [0,\infty)$ that $$ F_f(t) = \mu\{x\in X\,:\,f(x) \gt t\} = \lim\limits_{n\to\infty} \mu\{x \in X\,:\,g_n(x) \gt t\} = \lim\limits_{n\to\infty} F_{g_n}(t) $$ where the convergence is monotone in $n$. Again by the monotone convergence theorem we get $$\tag{2} \int_{0}^\infty \mu\{x\in X\,:\,f(x) \gt t\}\,dt = \lim\limits_{n\to\infty} \int_{0}^\infty \mu\{x \in X\,:\,g_n(x) \gt t\}\,dt. $$ Recalling the definitions of $E_{k}^n$ and $g_n$ we see that $$ \mu E_{k}^n = \begin{cases} \mu\{x\in X\,:\,g_n(x) \gt t\} & \text{if }1 \leq k \leq 4^n \text{ and } \frac{k-1}{2^n} \leq t \lt \frac{k}{2^n} \\ 0, & \text{otherwise,} \end{cases} $$ so that $$\tag{3} \int_{0}^\infty \mu\{x\in X\,:\,g_n(x) \gt t\}\,dt = \sum_{k=1}^{4^n} \frac{1}{2^n}\mu E_{k}^n = \int_{X} g_n\,d\mu. $$ Combining $(1)$ and $(2)$ using $(3)$ we get the desired formula $$ \int_X f\,d\mu = \int_{0}^\infty \mu\{x\in X\,:\,f(x) \gt t\}\,dt. $$

share|improve this answer
4  
...and that's the origin of the expression "Easy as 1,2,3". –  Asaf Karagila Aug 27 '12 at 22:52
    
Could you explain a bit more why the RHS is Riemann integrable? $F_f(t)$ can be infinite at some point. What if $\int_Xfd\mu=\infty$? –  Kaa1el Feb 24 at 19:22
add comment

Notations and Definitions Let $(X, \mu)$ be a measure space. Let $f\colon X \rightarrow [0, \infty]$ be a measurable function on $X$. We define a function $F_f\colon [0, \infty) \rightarrow [0, \infty]$ by

$F_f(t) = \mu(\{x \in X; f(x) > t\})$

We use the definitions of this question.

Lemma 1 Let $(X, \mu)$ be a measure space. Let $f\colon X \rightarrow [0, \infty]$ be a measurable function on $X$. Suppose $f$ is integrable on $X$. Then $F_f(t) < \infty$ for every $t > 0$.

Proof: Let $X_t = \{x \in X; f(x) > t\}$.

Then

$t\mu(X_t) \le \int_{X_t} f \, d\mu \le \int_X f \, d\mu < \infty$.

Since $t > 0$, $F_f(t) = \mu(X_t) < \infty$.

QED

Lemma 2 Let $[a, b]$ be a finite interval of the real line. Let $f\colon [a, b] \rightarrow (-\infty, \infty)$ be a monotone function on $[a, b]$. Then $f$ is Riemann integrable.

Proof: Without loss of generality, we can assume that $f$ is non-decreasing. Let $P\colon a = t_0 < t_1 <\cdots < t_{k-1} < t_k = b$ be a partition of $[a, b]$.

Then

$s_P = \sum_{i= 1}^k f(t_{i-1}) (t_i - t_{i-1})$.

$S_P = \sum_{i= 1}^k f(t_i) (t_i - t_{i-1})$.

Hence

$S_P - s_P = \sum_{i= 1}^k (f(t_i) - f(t_{i-1})) (t_i - t_{i-1})$

Let $\delta = \max\{t_i - t_{i-1}; i = 1,\dots,k\}$.

Then

$S_P - s_P \le \delta\sum_{i= 1}^k (f(t_i) - f(t_{i-1})) = \delta (f(b) - f(a))$

Hence $S_P - s_P$ can be arbitrarily small.

Let $\Phi$ be the set of partitions of $[a, b]$.

Let $s = \sup\{s_P; P \in \Phi\}$.

Let $S = \inf\{S_P; P \in \Phi\}$.

It is easy to see that $s \le S$. Since $s_P \le s \le S \le S_P$, $s = S$. Hence $f$ is Riemann integrable. QED

Lemma 3 Let $(X, \mu)$ be a measure space. Suppose $\mu(X) < \infty$. Let $f$ be a measurable function on $X$. Suppose $0 \le f(x) \le M$ for every $x \in X$, where $0 < M < \infty$. Let $P\colon 0 = t_0 < t_1 <\cdots < t_{k-1} < t_k = M$ be a partition of $[0, M]$

Let $m_i = \inf\{F_{f}(x); x \in [t_{i-1}, t_i]\}$.

Let $M_i = \sup\{F_{f}(x); x \in [t_{i-1}, t_i]\}$.

Let $s_P = \sum_{i= 1}^k m_i (t_i - t_{i-1})$.

Let $S_P = \sum_{i= 1}^k M_i (t_i - t_{i-1})$.

Let $A_i = \{x \in X; t_{i-1} < f(x) \le t_i\}$.

Then

$S_P = \sum_{i= 1}^k \mu(A_i)t_i$.

$s_P = \sum_{i= 1}^k \mu(A_i)t_{i-1}$.

Proof: Note that $F_f(t)$ is a monotone non-increasing function and $F_f(M) = 0$. Hence for $i = 1,2,\dots k$,

$M_i = F_f(t_{i-1})$

$m_i = F_f(t_i)$

Hence

$S_P = \sum_{i= 1}^k F_f(t_{i-1})(t_i - t_{i-1})$

$s_P = \sum_{i= 1}^k F_f(t_i)(t_i - t_{i-1})$

$S_P = F_f(t_0)t_0 + (F_f(t_0) - F_f(t_1))t_1 + \cdots + (F_f(t_{k-2}) - F_f(t_{k-1}))t_{k-1} + F_f(t_{k-1})t_k = (F_f(t_0) - F_f(t_1))t_1 + \cdots + (F_f(t_{k-2}) - F_f(t_{k-1}))t_{k-1} + (F_f(t_{k-1}) - F_f(t_k))t_k = \sum_{i= 1}^k (F_f(t_{i-1}) - F_f(t_i))t_i = \sum_{i= 1}^k \mu(A_i)t_i$

Similarly

$s_P = \sum_{i= 1}^k \mu(A_i)t_{i-1}$ QED

Lemma 4 Let $(X, \mu)$ be a measure space. Suppose $\mu(X) < \infty$. Let $f$ be a measurable function on $X$. Suppose $0 \le f(x) \le M$ for every $x \in X$, where $0 < M < \infty$.

Then $F_f$ is Riemann integrable and $\mathcal{R}(F_f, [0, M]) = \int_X f \, d\mu$.

Proof: This is clear from Lemma 3 and this result.

Lemma 5 Let $[a, \infty)$ be an infinite interval of the real line, where $-\infty < a < \infty$. Let $f_n\colon [a, \infty) \rightarrow [0, \infty)$ be a real valued function for $n = 1, 2, \dots$.Suppose $f_n \le f_{n+1} (n = 1,2,\dots)$. Suppose each $f_n$ is Riemann integrable on every finite interval $[x, y]$, where $a < x < y < \infty$. Let $f = \lim_n f_n$. Suppose $f$ is Riemann integrable on every finite interval $[x, y]$, where $a < x < y < \infty$.

Then $\lim_n \mathcal{R}(f_n, [a, \infty)) = \mathcal{R}(f, [a, \infty))$.

Proof: By this question, each $f_n$ is Lebesgue measurable and $\mathcal{R}(f_n, [a, \infty)) = \int_{a}^{\infty} f_n dt$. By Lebesgue monotone convergence theorem, $\lim_n \mathcal{R}(f_n, [a, \infty)) = \int_{a}^{\infty} f \, dt$.

By this question, $f$ is Lebesgue measurable and $\mathcal{R}(f, [a, \infty)) = \int_{a}^\infty f \, dt$.

This completes the proof. QED

Lemma 5.5 Let $[a, \infty)$ be an infinite interval of the real line, where $-\infty < a < \infty$. Let $f\colon [a, \infty) \rightarrow [0, \infty]$ be a non-increasing function. Suppose $f$ is extended Riemann integrable(see Notations and Definitions). Then $f(t)$ is finite for every $t > a$.

Proof: Suppose $f(t_0) = \infty$ for some $t_0 > a$. Since $f$ is non-increasing, $f(t) = \infty$ on $[a, t_0]$. Hence $\mathcal{R}(f, [a, t_0]) = \infty$. This is a contradiction. QED

Lemma 6 Let $(X, \mu)$ be a measure space. Let $f$ be a measurable function on $X$. Suppose $0 \le f(x) \le M$ for every $x \in X$, where $0 < M < \infty$.

Then $F_f$ is extended Riemann integrable if and only if $f$ is integrable. Moreover, if this is the case, $\mathcal{R}(F_f, [0, M]) = \int_X f \, d\mu$.

Proof: For every integer $n \ge 1$, let $X_n = \{x \in X; f(x) > 1/n\}$. Let $f_{n}$ be the function defined on $X$ which is equal to $f$ on $X_n$ and is equal to $0$ outside $X_n$. Then $f_{n} \le f_{n+1}$ for $n = 1, 2, \dots$ and $f = \lim_n f_{n}$.

Suppose $F_f$ is extended Riemann integrable. Since $F_f$ is non-increasing, by Lemma 5.5, $\mu(X_n) < \infty$ for $n = 1, 2, \dots$. Since $F_{f_{n}}$ is non-increasing and bounded, $F_{f_{n}}$ is Riemann integrable on $X$ by Lemma 2. Hence

$\mathcal{R}(F_{f_{n}}, [0, M]) = \int_{X_n} f_n d\mu = \int_X f_n d\mu$ by Lemma 4. Hence

$\lim_n \mathcal{R}(F_{f_{n}}, [0, M]) = \lim_n \int_X f_n d\mu$

We evaluate the both sides of this equation.

LHS = $\mathcal{R}(F_f, [0, \infty))$ by Lemma 5.

RHS = $\lim_n \int_X f \, d\mu$ by Lebesgue monotone convergence theorem. Hence

$\mathcal{R}(F_f, [0, \infty)) = \int_X f \, d\mu$

Hence $f$ is integrable on $X$.

Conversely suppose $f$ is integrable on $X$. By Lemma 1, $\mu(X_n) < \infty$ for $n = 1,2,\dots$. Hence $F_{f_n}$ is Riemann integrable on $[0, M]$ and $\mathcal{R}(F_{f_n}, [0, M]) = \int_{X_n} f_n d\mu$ by Lemma 4. By the similar argument as above, $\mathcal{R}(F_f, [0, M]) = \int_X f \, d\mu$. Hence $F_f$ is extended Riemann integrable on $[0, M]$. QED

Lemma 7 Let $(X, \mu)$ be a measure space. Let $f_n\colon X \rightarrow [0, \infty]$ be a real valued measurable function for $n = 1, 2, \dots$ Suppose $f_n \le f_{n+1}$ for $n = 1, 2, \dots$ Let $f = \lim_n f_n$. Then

$F_{f_n} \le F_{f_{n+1}}$ for $n = 1, 2, \dots$

and

$F_f = \lim_n F_{f_n}$ on $[0, \infty)$.

Proof: Since $f_n \le f_{n+1}$, $\{x \in X : f_n(x) > t\} \subset \{x \in X : f_{n+1}(x) > t\}$. Hence $F_{f_n} \le F_{f_{n+1}}$ for $n = 1, 2, \dots$ Since $\{x \in X : f(x) > t\} = \bigcup_n \{x \in X : f_n(x) > t\}$, $F_f = \lim_n F_{f_n}$ on $[0, \infty)$. QED

Proposition Let $(X, \mu)$ be a measure space. Let $f\colon X \rightarrow [0, \infty]$ be a measurable function on $X$. Then $F_f$ is extended Riemann integrable on $[0, \infty)$ if and only if $f$ is integrable on $X$. If this is the case, $\mathcal{R}(F_f, [0, \infty)) = \int_X f \, d\mu$.

Proof: Suppose $F_f$ is extended Riemann integrable on $[0, \infty)$. Let $f_n(x) = \min\{f(x), n\}(n = 1,2,\dots)$. Then $f_n$ is bounded and $f_n \le f_{n+1} (n = 1,2,\dots)$ and $f = \lim_n f_n$. Since $F_f$ is extended Riemann integrable, each $F_{f_n}$ is extended Riemann integrable.

By Lemma 6, $f_n$ is integrable and $\mathcal{R}(F_{f_n}, [0, n]) = \int_X f_n d\mu$.

Since $\mathcal{R}(F_{f_n}, [0, n]) = \mathcal{R}(F_{f_n}, [0, \infty))$,

$\lim_n \mathcal{R}(F_{f_n}, [0, \infty)) = \lim_n \int_X f_n d\mu$.

We evaluate the both sides of this equation.

By Lemma 7, $F_{f_n} \le F_{f_{n+1}}$ for $n = 1, 2, \dots$ and $F_f = \lim_n F_{f_n}$.

Hence, by Lemma 5, $\lim_n \mathcal{R}(F_{f_n}, [0, \infty)) = \mathcal{R}(F_f, [0, \infty))$.

On the other hand, by Lebesgue monotone convergence theorem,

$\lim_n \int_X f_n d\mu = \int_X f \, d\mu$

Hence $\mathcal{R}(F_f, [0, \infty)) = \int_X f \, d\mu$.

Conversely suppose $f$ is integrable on $X$. Let $f_n(x) = \min\{f(x), n\}(n = 1,2,\dots)$. Then $f_n$ is bounded and $f_n \le f_{n+1} (n = 1,2,\dots)$ and $f = \lim_n f_n$. Since each $f_n$ is integrable, by Lemma 6, $F_{f_n}$ is extended Riemann integrable and $\mathcal{R}(F_{f_n}, [0, n]) = \int_X f_n d\mu$.

By Lemma 7, $F_{f_n} \le F_{f_{n+1}}$ for $n = 1, 2, \dots$ and $F_f = \lim_n F_{f_n}$. Hence, by Lemma 5, $\lim_n \mathcal{R}(F_{f_n}, [0, \infty)) = \mathcal{R}(F_f, [0, \infty))$. On the other hand, by Lebesgue monotone convergence theorem,

$\lim_n \int_X f_n d\mu = \int_X f \, d\mu$

Hence $\mathcal{R}(F_f, [0, \infty)) = \int_X f \, d\mu$. QED

share|improve this answer
    
@MichaelHardy please edit correctly. –  Makoto Kato Sep 25 '12 at 16:49
    
Please leave a comment explaining the downvote so that I can improve my answer. –  Makoto Kato Sep 25 '12 at 17:32
    
What is it that you object to about the edits I did? –  Michael Hardy Sep 26 '12 at 0:11
    
BTW, I agree with you that those who down-vote postings should explain their objections. –  Michael Hardy Sep 26 '12 at 1:45
    
@MichaelHardy You did not edit "Notations and Definitions" correctly. Please see the edit history. Anyway, thanks for the other edits. –  Makoto Kato Sep 26 '12 at 2:24
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.