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Let $f$ be Riemann integrable function and $g$ be Lebesgue integrable function on $[0,1]$. If $\int_{0}^{1} |f-g| = 0$, then what we can say about $g$ i.e. whether $g$ will be Riemann integrable or not?

I feel $g$ will be Riemann integrable, because if $g$ is not Riemann integrable that means set of discontinuities of $g$ must have non zero measure. But from $\int_{0}^{1} |f-g| = 0$, we can deduce that $f$ and $g$ are equal almost everywhere. I want to verify whether I am thinking in right direction or not. Please help me to figure out this problem, thanks in advance.

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No, this is not right. Consider $f=0$ and $g=1_\mathbb{Q}$. In particular, Riemann integrability requires that for almost all $x$, $g$ is continuous in $x$. This is not the same as being equal to a continuous function almost everywhere.

The difference is clear here: $g$ is equal to the continuous function $0$ a.e., but it is actually continuous nowhere.

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  • $\begingroup$ Ok thanks, I got it. Actually I did not try to find a counterexample but instead, I tried to prove. $\endgroup$ Commented Jun 27, 2020 at 17:42
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If $f\colon[0,1]\longrightarrow\Bbb R$ is the null function and if $g\colon[0,1]\longrightarrow\Bbb R$ is $1$ on the rationals and $0$ otherwise, then$$\int_0^1|f-g|=0,$$but $g$ is not Riemann-integrable.

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  • $\begingroup$ I got it, nice counterexample. $\endgroup$ Commented Jun 27, 2020 at 17:43

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