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


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


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.


1 Answer 1


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!


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