Does Lebesgue integrability imply improper Riemann integrability for positive, a.e. continuous functions?

Let $E \subseteq \mathbb{R}^n$ be a open subset and let $f:E \to \mathbb{R}$ be positive (not nessesarily bounded) and a.e. continuous and suppose that the Lebesgue-integral of $f$ over $E$ exists (i.e. is finite). Is it true, that then $f$ is improper Riemann integrable over $E$?

The improper Riemann integral is defined as follows:

For every $M > 0$ let $g_M = \min\{f,M\}$. Then $f$ is improper Riemann integrable iff $\lim_{M \to \infty} \int_E g_M(x)dx$ exists, where the integral is the Riemann integral.

• Yes, of course. Thank you – Andrei Kh May 11 '18 at 16:32
• Is $E$ Jordan measurable? Notice there are open sets which aren't Jordan measurable. – user251257 May 11 '18 at 16:44
• Yes, I forgot to mention it. But thanks for the counterexample in this case! – Andrei Kh May 11 '18 at 17:33

If $M_n\to \infty$ then $g_{M_n} \to f$ increasingly. So $\int g_{M_n} \to \int f$ (even if the latter is infinite). So for positive functions, $f$ is R integrable improperly if and only if $f$ is L integrable. You can do any other truncations for $f$ positive, provided they converge to $f$ increasingly.
If $E$ is Jordan measurable, then the statement is true due to one of many convergence theorem of Lebesgue integral.
If $E$ is not Jordan measurable, then it is wrong. Let $E$ be the complement of the Fat Cantor set in $(0,1)$. Then, $E$ is open and not Jordan measurable. In particular, it's indicator $\chi_E$ isn't Riemann integrable but Lebesgue integrable.