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If $f$ is a Riemann integrable function on $[0,2\pi]$ such that $f(x)=0$ whenever $f$ is continuous at $x$, then: \begin{equation} \|f\|=\left(\frac{1}{2\pi}\int_0^{2\pi}|f(x)|^2dx \right)^{1/2}=0 \end{equation}

How do I go about proving this without any measure theory? I've thought about using the fact that the Cesaro mean of the Fourier series converges to 0 at all points of continuity but it leads to nowhere.

C. Webb
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  • The hypothesis just implies that $f=0$ a.e. https://math.stackexchange.com/questions/238139/riemann-integrable-and-continuous-almost-everywhere – PNDas Jan 27 '23 at 06:43
  • The proof that I was lead to required the notion of a measure. I have yet to cover any measure theory as of yet, so is there perhaps a way without invoking the Lebesgue criterion for Riemann integrability? – C. Webb Jan 27 '23 at 07:08
  • It requires the notion of a measure, but not really. It only requires that you know the definition of a measure zero set: We say $E$ has measure zero if for every $\epsilon >0$ there is a countable collection of disjoint closed intervals such that $\cup_{n\geq 1}[a_n,b_n]\supset E$ and $\sum_{n\geq 1}b_n-a_n <\epsilon$. The proof of Lebesuge criterion also does not use anything fancy really. – Andrew Jan 27 '23 at 07:27

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Hints.

Step I. If $f:[a,b]\to\mathbb R$ is Riemann integrable, then $f$ is continuous in a dense subset of $[a,b]$.

Step II. If $f:[a,b]\to\mathbb R$ is Riemann integrable, then so it $f^2$.

Step III. If $g:[a,b]\to\mathbb R$ is Riemann integrable, and vanishes in a dense subset of $[a,b]$, then $\int_a^b g=0$.

Yiorgos S. Smyrlis
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