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Given a measure space $\Omega$.

Consider plain functions $f$.

Define the plain integral as usual: $$f\geq0:\quad\int f\mathrm{d}\mu:=\sup_{s\leq f}\int s\mathrm{d}\mu$$

Do vanishing integrals imply measurability: $$\int|f-s_n|\mathrm{d}\mu\to0\implies s_{n_k}\to f\implies f\in\mathcal{M}$$ (Be aware that the usual proof heavily depends upon measurability!)

Have a guess! ;)

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1 Answers1

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No, they don't!

Consider a Vitali set $V\subseteq[0,1]$.

Then one has trivially: $$\int|\chi_V-0|\mathrm{d}\mu\equiv\mu_*(V)=0$$ However, it is not even measurable!

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