# 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 $\mu$ be the measure, and define for $t >0$,

and

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

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

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.

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

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

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

Moreover, for every $t \in [0,\infty)$, we have

so, using the notation in the question, we have for all $t \in [0,\infty)$ that

where the convergence is monotone in $n$. Again by the monotone convergence theorem we get

Recalling the definitions of $E_{k}^n$ and $g_n$ we see that

so that

Combining $(1)$ and $(2)$ using $(3)$ we get the desired formula