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.
Best Answer
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:
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 $$ \int_X f\,d\mu = \int_{0}^\infty \mu\{x\in X\,:\,f(x) \gt t\}\,dt. $$