Definition of the Lebesgue integral

Question

I found a new (for me) definition for the Lebesgue integral by the Riemann integral:

$\int f d \mu = \int_0^{\infty} \mu(f^{-1}((t,\infty]))dt \in [0,\infty]$.

Does this definition makes sense for you, did you ever seen it? I would be very happy for some explanations :)

Regards from Berlin!

Answer 1

Let $f$ be a non-negative measurable function. Then

$$ f(x)=\int_0^{f(x)}dt=\int_0^{+\infty}\chi_{\{f>t\}}(x)dt $$ where $\{f>t\}=\{x\in\mathbb R^N, f(x)>t\}$. As such, using Fubini's theorem,

\begin{align} \int_{\mathbb R^N} f(x)dx &= \int_{\mathbb R^N}\left(\int_0^{+\infty}\chi_{\{f>t\}}(x)dt\right)dx \\ &=\int_0^{+\infty}\left(\int_{\mathbb R^N}\chi_{\{f>t\}}(x)dx\right)dt\\ &=\int_0^{+\infty}\mu(\{f>t\})dt. \end{align}

We can make sense of this formula by giving it a geometric interpretation. It is related to the Riemann integral in the following sense: we are calculating the area under the graph of $f$ by summing slices of it. Heuristically, we can say that $$ \mu(\{t<f<t+h\})\simeq h\,\mu(\{t<f\}). $$

It is not exactly the same thing as the Riemann integral though, where the approximation is done with vertical slices, $$ \int_a^b f=\lim_{n\to\infty} \sum_{k=0}^n \frac{f(a+k(b-a)/n)}{n}, $$ whereas with the Lebesgue integral, the slices are horizontal. A direct consequence of this difference is that the Lebesgue integral is easy to define for functions $f:X\to \mathbb R$ (or $\mathbb R^N$), where $X$ is somewhat arbitrary, but the Riemann integral is well suited for functions $f:\mathbb R\to X$, where, once again, $X$ is somewhat arbitrary.

Answer 2

This is for nonnegative measurable functions $f$.

Yes, you can find it in many textbooks. For example in probability textbooks, computing the expectation of a random variable knowing the cumulative distribution function of it.

Also, for some instructional purposes, we may avoid the Lebesgue integral by writing $$ \int f d \mu = \int_0^{\infty} \mu(f^{-1}((t,\infty]))dt $$ since the integral on the right is (improperly) Riemann integrable.

Answer 3

Ok, thats cool! Let us take a look at an example: $\int_{B_1(0)} \vert x \vert^s d \mu(x)$ where $\mu$ denotes the n-dimensional Lebesgue measure and $B_1(0)$ the unit ball and $s \in \mathbb{R}$.

Is it possible here to work with the definition above?