Consider the following function:
$$ f(x)=\begin{cases} 0 & x\text{ is a rational number}, \\ 1 & x\text{ is an irrational number}. \end{cases} \ $$
Show that the function $f(x)$ is not integrable over any interval $[a,b]$.
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2try to prove the function is not continuos – y_andoni Feb 10 '21 at 18:50
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2@y_andoni There exist integrable non continuous functions – Raffaele Feb 10 '21 at 18:51
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1https://math.stackexchange.com/q/1079172/ – Feb 10 '21 at 18:52
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1https://math.stackexchange.com/q/1550808/ – Feb 10 '21 at 18:52
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2Prove that inf sum is zero and sup sum is $b-a$ so they don't converge https://en.wikipedia.org/wiki/Dirichlet_function – Raffaele Feb 10 '21 at 18:53
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2@Raffaele well its not riemann integrable but its integrable as a Lebesgue integral . Since the OP asked to show the function is not integrable i assumed he meant Riemann-integrable. And the function is also non-monotonous so you are only left to check if it is continous or not. – y_andoni Feb 10 '21 at 18:55
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3@y_andoni A function only needs to be continuous almost everywhere to be Riemann integrable. – Sandejo Feb 10 '21 at 19:01
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1@y_andoni, Yes, the question assumes Riemann integration. – Adnan Ali Feb 10 '21 at 19:02
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1Hint: use the definition of "Riemann-integrable". – Daniel Hast Feb 10 '21 at 19:03
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1@y_andoni The given function is not continuous on the set of rationals. – TheSilverDoe Feb 11 '21 at 08:03
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The function does not have left-hand and right-hand limits at each point, hence it is not regulated, hence it is not Riemann-integrable.
TheSilverDoe
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