User:CaptchaSamurai/sandbox/HK-IntegrabilityOfCertainFunctions

Integrability of certain functions

Function	Riemann	Improper Riemann	Lebesgue	Henstock-Kurzweil
$${\begin{aligned}f\colon [0,1]&\to \mathbb {R} \\x&\mapsto {\begin{cases}1/{\sqrt {x}},&x\in (0,1],\\0,&x=0\end{cases}}\end{aligned}}$$	No. Function $f$ is unbounded and only bounded functions are Riemann integrable.	Yes. We have$$\lim _{a\to 0+}\int _{a}^{1}f=\lim _{a\to 0+}2-{\sqrt {a}}=2.$$	Yes. By the monotone convergence theorem$${\begin{aligned}\int f&=\int \lim _{n}f\cdot \chi _{[1/n,1]}\\&=\lim _{n}\int f\cdot \chi _{[1/n,1]}.\end{aligned}}$$The last integral is Riemann integrable for each $n$ and the limit converges.	Yes. Function $f$ is a derivative of $x\mapsto 2{\sqrt {x}}.$
The indicator function of rationals on the unit interval, that is$${\begin{aligned}\chi _{\mathbb {Q} \cap [0,1]}\colon [0,1]&\to \mathbb {R} \\x&\mapsto {\begin{cases}1,&x\in [0,1]\cap \mathbb {Q} ,\\0,&x=0.\end{cases}}\end{aligned}}$$.	No. The set of discontinuities has a positive measure (the Lebesgue-Vitali theorem of characterization of the Riemann integrable functions).		Yes. This function differs from a constant function on a measure zero set.	Yes. By a direct proof or the fact that Lebesgue-integrable implies H-K integrable.
The derivative of Volterra's function on $[-1,1]$, namely $v'\colon [0,1]\to \mathbb {R}$, where$${\begin{aligned}v\colon [0,1]&\to \mathbb {R} \\x&\mapsto {\begin{cases}x^{2}\sin(1/x),&x\in (0,1],\\0,&x=0.\end{cases}}\end{aligned}}$$	No. The set of discontinuities has a positive measure.		Yes. The function $v$ is absolutely continuous. This implies that $v'$ is Lebesgue integrable.	Yes. It's a derivative.
Let $g=G'$, where $G(x)=x^{2}\sin({\frac {1}{x^{2}}})$ for $x\in (0,1]$ and $G(0)=0$. Explicity $$g(x)=2x\sin({\frac {1}{x^{2}}})-{\frac {2}{x}}\cos({\frac {1}{x^{2}}}),$$whenever $x\in (0,1]$, and $g(0)=0.$	No. Function is unbounded.	Yes. ${\begin{aligned}\lim _{a\to 0+}\int _{a}^{1}G'&=\lim _{a\to 0+}G(1)-G(a)\\&=G(1)<\infty \end{aligned}}$	No. Integral of $|g|$ is unbounded.	Yes. It's a derivative.
${\begin{aligned}h\colon [0,\infty )&\to \mathbb {R} \\x&\mapsto \sum _{n=1}^{\infty }\chi _{[n,n+1)}{\frac {(-1)^{n}}{n}}\end{aligned}}$	Not defined (domain is not compact).	Yes. $\lim _{n\to \infty }\int _{1}^{n}h=\sum _{n=1}^{\infty }{\frac {(-1)^{n}}{n}}=\ln 2$.	No. The integral of $|h|$ is unbounded.	Yes.
$${\begin{aligned}a\colon [0,\infty )&\to \mathbb {R} \\x&\mapsto {\begin{cases}\chi _{\mathbb {Q} }(x)&x\in [0,1),\\h(x)&x\in [1,\infty )\end{cases}}\end{aligned}}$$	Not defined (domain is not compact).	No. The set of discontinuities has a positive measure.	No. The integral of $|a|$ is unbounded.	Yes.
Indicator function of a non-measurable set, like some instance of a Vitali set.	No. We can only integrate functions that are measurable.