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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.

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    This definition seems to be a bit more intuitive in my opinion. Thank you for shareing.2015-11-02

2 Answers 2

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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. $

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    Could you explain a bit more why the RHS is Riemann integrable? $F_f(t)$ can be infinite at some point. What if $\int_Xfd\mu=\infty$?2014-02-24
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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

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    @MichaelHardy You did not edit "Notations and Definitions" correctly. Please see the edit history. Anyway, thanks for the other edits.2012-09-26