if $f \in L^1,$ is there a Riemann integrable $g$ s.t. $f=g$ almost everywhere?

Question

I'm currently studying the basic properties of $L^p$ spaces, and the following question popped to my mind.

If $f \in L^1(\mathbb{R}),$ does it hold that there exists a Riemann integrable function $g$ such that $f=g$ almost everywhere? I can't immediately come up with neither a proof nor a counterexample.

Answer 1

The following example comes from Carothers. Let $G$ be an open subset of $[0,1]$ containing the rationals in $[0,1]$ such that the outer measure of $G$, $\lambda^*(G)$, satisfies $\lambda^*(G)<1/2$. If $f=1_G$, then $f$ is not Riemann integrable, and also cannot be equal a.e. to ANY Riemann integrable function. See if you can show this! A key result in doing so is that for a bounded function $f$, $f$ is Riemann integrable if and only if $f$ is continuous at almost every point.