Two definitions of Lebesgue integration

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 μ be the measure, and define for t>0,
Sf(t)={xX:f(x)>t},
and
Ff(t)=μ(Sf(t)).

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

Xfdμ=0Ff(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.

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:

Let (X,Σ,μ) be a measure space and let f:X[0,) be a measurable function. Then for the usual Lebesgue integral Xfdμ of f over (X,Σ,μ) we have the identity
Xfdμ=0μ{xX:f(x)>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,), and thus its Riemann integral exists in the extended sense and the Riemann and Lebesgue integrals coincide for it.

To prove this, set Enk={xX:f(x)>k2n} and put
gn=12n4nk=1[Enk]
where [A] denotes the characteristic function of A. Note that we have 0f(x)gn(x)<2n whenever 0f(x)<2n, so that gn(x)f(x) for every xX. Also, gn(x)gn+1(x) for all x, so the sequence (gn)nN increases to f pointwise everywhere. By the monotone convergence theorem we conclude that
Xfdμ=lim
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.

Attribution
Source : Link , Question Author : Doubting Thomas , Answer Author : t.b.

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