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)={x∈X: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{x∈X: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={x∈X:f(x)>k2n} and put

gn=12n4n∑k=1[Enk]

where [A] denotes the characteristic function of A. Note that we have 0≤f(x)−gn(x)<2−n whenever 0≤f(x)<2n, so that gn(x)→f(x) for every x∈X. Also, gn(x)≤gn+1(x) for all x, so the sequence (gn)n∈N 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.*