Lebesgue Integral of Simple Functions

measure-theoryreal-analysis

I am mainly confused by the notation. An excerpt from Stein and Shakarchi.

"If $E$ is a measurable subset of $\mathbb{R}^d$ with finite measure, then $\varphi(x)\chi_E(x)$ is also a simple function, and we define

$$\int_E\varphi(x)\,dx=\int\varphi(x)\chi_E(x)\,dx.$$

To emphasize the choice of the Lebesgue measure $m$ in the definition of the integral, one sometimes writes

$$\int_{\mathbb{R}^d}\varphi(x)\,dm(x)."$$

My Question:

  1. The Lebesgue measure partitions the $range$ of the function, so why are we still taking the integral with respect to $x$? Is this because we looking at the measurable set $E$, which are a collection of $x$ such that it falls into a particular partition in the range?

  2. What does $dm(x)$ mean in the second statement. Are we integrating with respect to the measure? How are they equivalent?

Reference:
Real Analysis: Measure Theory, Integration, and Hilbert Spaces. Elias M. Stein, Rami Shakarchi. Princeton University Press, 2009.

Best Answer

The choice of writing $dx$ in the first two integrals is unfortunate. It is a nearly universal convention that when you write $dx$ as the differential of an integral, you mean Riemann integration, and when you write $dm$ or $d\mu$ as the differential, you mean Lebesgue integration. So the answer to your first question is that we're not taking the integral with respect to $x;$ in context, your book is definitely talking about Lebesgue integration.

You are integrating with respect to the measure $m$ in the second statement. It's another unfortunate choice of notation: most authors would have written $$\int_{\mathbb{R}^d}\varphi(x)\,dm\quad\text{or}\quad \int_{\mathbb{R}^d}\varphi(x)\,d\mu$$ for the second integral, depending on whether they were talking about the Lebesgue measure, or a more general measure.

There is a theorem that says if a function $f$ is Riemann integrable, then the Riemann integral over a region is equal to the Lebesgue integral over the same region. And that only makes sense. Counting up your area in two different ways shouldn't give you two different results!

Related Question