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Hello fellow Mathematics enthusiasts. I was hoping someone could help me with the following problem from Terry Tao's Introduction to Measure Theory:

Let $[a,b]$ be an interval, and let $f,g:[a,b] \to \mathbb{R}$ be Riemann integrable. Establish the following statement.

(Indicator) If $E$ is a Jordan measurable of $[a,b]$, then the indicator function $1_E: [a,b] \to \mathbb{R}$ (defined by setting $1_E(x) :=1$ when $x \in E$ and $1_E(x) :=0$ otherwise.) is Riemann integrable, and $\int_{a}^{b}1_E(x) dx = m(E)$.

In this problem, the notion of Jordan measure is being used. As a quick refresher, the Jordan inner measure $m_{*,(J)}(E) := \sup_{A \subset E, A \quad \text{elementary}} m(A)$ and the Jordan outer measure $m^{*,(J)}(E) := \inf_{B \supset E, B \quad \text{elementary}} m(B)$. Whenever $m_{*,(J)}=m^{*,(J)}$, then we say that $E$ is Jordan measurable and call $m(E)$ the Jordan measure of $E$.

This is the third part of an exercise that asks the reader to establish some basic properties of the Riemann integral: linearity and monotonicity. I have done the previous two, but do not know how to start this one. Any help will be greatly appreciated, thanks in advance.

Jamil_V
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1 Answers1

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First of all $A \subset B\,$ iff $\,1_A \le 1_B$ .

Then, since $E$ is measurable, for every $\varepsilon>0$ there exist two elementary sets, say $A$ and $B$, such that $$A \subset E \subset B$$ and $$m(B\setminus A)=m(B)-m(A) \le \varepsilon$$ (see exercise 1.1.5 (2)).

Now (see exercise 1.1.21 (3) and the Darboux definition) $$m(A) \le \underline{\int_a^b} 1_E(x)\,dx$$ and $$m(B) \ge \overline{\int_a^b} 1_E(x)\,dx$$ so $$\overline{\int_a^b} 1_E(x)\,dx-\underline{\int_a^b} 1_E(x)\,dx \le m(B)-m(A) \le \varepsilon$$ The arbitrariness of $\varepsilon$ gives the integrability of $1_E$ .

Finally $$m(A) \le m(E) \le m(B)$$ and $$m(A) \le \int_a^b 1_E(x)\,dx \le m(B)$$ so (see the simple fact in my answer) $$\int_a^b 1_E(x)\,dx=m(E)$$

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Tony Piccolo
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