Exercise 1.3.17 (Lebesgue integral supercedes Riemann-Darboux integral)

Question

Let $f:[a,b]	o[0,+\infty]$ be a Riemann integrable function, extended to the domain $\mathbb{R}$ by declaring it to be zero outside of $[a,b]$. Show that
    $$\int_{\mathbb{R}^d} f = \mathrm{Riemann}\int_{[a,b]} f$$

Elu Thingol · Accepted Answer

It is possible to demonstrate, using Darboux definition, that both integrals are equal by definition:
$$\sup\left\{ \mathrm{Simp}\int_{\mathbb{R}^{d}} h : \begin{array}{l} h 	ext{ simple}\ 0 \leq h \leq f \end{array} ight\} = \sup\left\{ \mathrm{pc}\int_{[a,b]} g : \begin{array}{l} g 	ext{ piecewise constant}\ 0 \leq g \leq f \end{array} ight\}.$$
However, a shortcut is to use the area under the curve of $f$. On one hand, by \href{./exercise_1-1-25}{Exercise 1.1.25} $f$ being Riemann integrable implies that the area under the curve has the Jordan measure equal to the Riemann integral of $f$. Using \href{exercise_1-2-8}{Exercise 1.2.8} we conclude
$$\mathrm{Riemann}\int_{[a,b]} f = m\left(\left\{(x,t) \in [a,b] 	imes [0,+\infty]: 0 \leq t \leq f(x) ight\} ight) $$
On the other hand, by \href{./exercise_1-3-13}{Exercise 1.3.13} the Lebesgue integral of $f$ is also equal to the area under the curve of $f$:
$$\int_{\mathbb{R}^d} f = \int_{\mathbb{R}^d} f \cdot 1_{[a,b]}
= m\left(\left\{(x,t) \in \mathbb{R} 	imes \mathbb{R}: 0 \leq t \leq f(x)\cdot 1_{[a,b]} ight\} ight)
= m\left(\left\{(x,t) \in [a,b] 	imes [0,+\infty]: 0 \leq t \leq f(x) ight\} ight).$$
Thus, both agree by transitivity.

Exercise 1.3.17 (Lebesgue integral supercedes Riemann-Darboux integral)

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Exercise 1.3.17 (Lebesgue integral supercedes Riemann-Darboux integral)

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