# When is a Lebesgue integrable function a Riemann integrable function?

When is a Lebesgue integrable function a Riemann integrable function ?

And if we have $$f\in \mathcal{L}^1([0,1],\lambda)$$, does it implies that $$f$$ is Riemann integrable, and why ?

No, it doesn't.

It was proven by Lebesgue that a bounded function $$f$$ is Riemann integrable precisely when the set of points where $$f$$ is discontinuous has measure zero.

• So in what case we can go from lebesgue inegral to reimann integral ? When do we have $\int_{a}^{b}f(x)dx=\int_{[a,b]}fd\lambda$ for a Lebesgue integrable function $f$ Commented Jan 15, 2019 at 18:46
• @AnasBOUALII Precisely when - as this answer says - the set of discontinuities of $f$ has measure zero. Is there something unsatisfying about this? Commented Jan 15, 2019 at 18:47
• But does every Lebesgue integrable function is a.e continious and bounded ? Commented Jan 15, 2019 at 18:49
• No, of course not. In your setup, the function $1_{\mathbb Q}$, the characteristic of the rationals, is everywhere discontinuous. And $f(x)=\tfrac1{\sqrt x}\,1_{\mathbb Q}$ is unbounded, everywhere discontinuous, and still in $L^1[0,1]$. Commented Jan 15, 2019 at 18:52
• @bof: yes. My example was about a Lebesgue measurable function which is not Riemann integrable. Commented Jan 15, 2019 at 19:05

This is the Riemann-Lebesgue Theorem, which says that $$f$$ is Riemann integrable if and only if the set of discontinuity points is of measure zero.

Note that $$1_{\mathbb{Q}}$$ (the indicator of the rationals) is Lebesgue-integrable but not Riemann integrable. On $$[0,1]$$ it is almost everywhere bounded but not continuous. Here, the set of discontinuity points is of measure $$1$$. So we cannot say anything about continuity and Lebesgue integrability (except that every continuous function on a set of finite measure is Lebesgue-integrable).