Function
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Riemann
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Improper Riemann
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Lebesgue
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Henstock-Kurzweil
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![{\displaystyle {\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e2c5ad747e7892964ee2790d02651e6ba64adb61)
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No.
Function is unbounded and only bounded functions are Riemann integrable.
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Yes. We have![{\displaystyle \lim _{a\to 0+}\int _{a}^{1}f=\lim _{a\to 0+}2-{\sqrt {a}}=2.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1bf11c438b0836d137801a5f2957540c8fda5350)
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Yes. By the monotone convergence theorem![{\displaystyle {\begin{aligned}\int f&=\int \lim _{n}f\cdot \chi _{[1/n,1]}\\&=\lim _{n}\int f\cdot \chi _{[1/n,1]}.\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2927c7b7e7fc6b9404a64ae475f1188c2fbeb778) The last integral is Riemann integrable for each and the limit converges.
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Yes.
Function is a derivative of
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The indicator function of rationals on the unit interval, that is![{\displaystyle {\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4e4d9cb65581efa650c6cd97f22f3ffd15229b88) .
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No.
The set of discontinuities has a positive measure (the Lebesgue-Vitali theorem of characterization of the Riemann integrable functions).
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Yes.
This function differs from a constant function on a measure zero set.
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Yes.
By a direct proof or the fact that Lebesgue-integrable implies H-K integrable.
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The derivative of Volterra's function[1] on , namely , where![{\displaystyle {\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f6e13bf2abfb14631b9b8a2d96722994b6e193cc)
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No.
The set of discontinuities has a positive measure.
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Yes.
The function is absolutely continuous. This implies that is Lebesgue integrable.
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Yes.
It's a derivative.
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Let , where for and . Explicity ![{\displaystyle g(x)=2x\sin({\frac {1}{x^{2}}})-{\frac {2}{x}}\cos({\frac {1}{x^{2}}}),}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3d3a63b4683c572d82cd2c7610f19063a5c7aad4) whenever , and
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No.
Function is unbounded.
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Yes.
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No.
Integral of is unbounded.
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Yes.
It's a derivative.
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Not defined (domain is not compact).
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Yes.
.
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No.
The integral of is unbounded.
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Yes.
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![{\displaystyle {\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/eaf6ef49a2eb9ba2623347fec7a2d4573721c15d)
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Not defined (domain is not compact).
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No.
The set of discontinuities has a positive measure.
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No.
The integral of is unbounded.
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Yes.
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Indicator function of a non-measurable set, like some instance of a Vitali set.
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No.
We can only integrate functions that are measurable.[2]
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