Let $f:[0,1] \to \mathbb{R}$ a function such that the set of it's points of discontinuities have measure $0$ and such that the improper Riemann integral $\int_{0}^{1} f(x) dx$ exists. It is true that $f$ is Lebesgue integrable in $[0,1]$ ? My guess is that this is true, but I don't know how to give an argument.
Relation between Riemann integral and Lebesgue integral
analysislebesgue-integrallebesgue-measuremeasure-theoryreal-analysis
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Best Answer
Here is an counter-example:
Define $f:[0,1] \to \mathbb{R}$ such that $f(x)=n(-1)^n$ if $\frac{1}{n+1} <x\leq \frac{1}{n}$ and $f(x)=0$ if $x=0$. Then $f$ has an countable points of discontinuities, and then the set of discontinuities of $f$ has measure $0$. Now we have: $$\int_{0}^{1} f(x) dx=\sum_{n=1}^{\infty} \frac{(-1)^{n}}{n+1}=\log2. $$
Showing that $f$ has improper Riemann integral. But
$$\int_{0}^{1}|f(x)|dx=\sum_{n=1}^{\infty} \frac{1}{n+1}=+\infty.$$
Concluding that $f$ is not Lebesgue integrable.