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Definition/Fact: We say a bounded subset $E \subset \mathbb{R}^d$ is Jordan measurable if its characteristic function $\chi_E$ is Riemann integrable. Questions: For simplicity we only consider subsets of $[0, 1]$. Show the following properties. (a) $\emptyset$ and $[0, 1]$ are Jordan measurable. (b) If $E \subset [0, 1]$ is Jordan measurable, then so is $E^c = [0, 1] \setminus E$. (c) If $E, F \subset [0, 1]$ are Jordan measurable, then so are $E \cup F$ and $E \cap F$. (d) Give an example where $E_1 \subset E_2 \subset E_3 \subset \cdots$ in $[0, 1]$ are all Jordan measurable, but $E = \bigcup_{n=1}^{\infty} E_n$ is not Jordan measurable. \textit{Hint: Dirichlet function is not Riemann integrable.} (This says that Jordan measurable sets do NOT form a $\sigma$-algebra.)

          Definition/Fact: We say a bounded subset $E \subset \mathbb{R}^d$ is Jordan measurable if its characteristic function $\chi_E$ is Riemann integrable.
Questions: For simplicity we only consider subsets of $[0, 1]$. Show the following properties.
(a) $\emptyset$ and $[0, 1]$ are Jordan measurable.
(b) If $E \subset [0, 1]$ is Jordan measurable, then so is $E^c = [0, 1] \setminus E$.
(c) If $E, F \subset [0, 1]$ are Jordan measurable, then so are $E \cup F$ and $E \cap F$.
(d) Give an example where $E_1 \subset E_2 \subset E_3 \subset \cdots$ in $[0, 1]$ are all Jordan measurable, but $E = \bigcup_{n=1}^{\infty} E_n$ is not Jordan measurable.
\textit{Hint: Dirichlet function is not Riemann integrable.}
(This says that Jordan measurable sets do NOT form a $\sigma$-algebra.)

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Definition/Fact: We say a bounded subset E ⊂ℝ^d is Jordan measurable if its characteristic function is Riemann integrable.
Questions: For simplicity we only consider subsets of [0, 1]. Show the following properties.
(a) ∅ and [0, 1] are Jordan measurable.
(b) If E ⊂ [0, 1] is Jordan measurable, then so is E^c = [0, 1] ∖ E.
(c) If E, F ⊂ [0, 1] are Jordan measurable, then so are E ∪ F and E ∩ F.
(d) Give an example where E1 ⊂ E2 ⊂ E3 ⊂⋯ in [0, 1] are all Jordan measurable, but E = ⋃n=1^∞ En is not Jordan measurable.
Hint: Dirichlet function is not Riemann integrable.
(This says that Jordan measurable sets do NOT form a σ-algebra.)

Added by Jill C.

Calculus: Early Transcendentals

James Stewart 8th Edition

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Definition/Fact: We say a bounded subset E âŠ† â„ is Jordan measurable if its characteristic function Xâ‚‘ is Riemann integrable. Questions: For simplicity, we only consider subsets of [0,1]. Show the following properties: a) 0 and 1 are Jordan measurable. b) If E âŠ† [0,1] is Jordan measurable, then so is Eâ‚– = [0,1] \ E. c) If E and F are Jordan measurable subsets of [0,1], then so are E âˆª F and E âˆ© F. d) Give an example where E, C, and Eâ‚ƒ are all Jordan measurable in [0,1], but E = E âˆª Eâ‚ƒ is not Jordan measurable. (Hint: The Dirichlet function is not Riemann integrable.) This says that Jordan measurable sets do NOT form a Ïƒ-algebra.

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