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Let $f:\mathbb{R}\to [0,\infty)$ be a measurable function and $f\in L^1$. Prove that for every $\varepsilon >0$ it exists $\delta >0$ such that $E\in \mathcal{M},\: m(E)<\delta \Rightarrow \int_Ef dμ<\varepsilon $

$f$ is non-negative, so $\int_Ef dμ=\sup \{\int_R\phi X_E \:dμ:0\leq\phi\leq f, \phi \:\text{simple and }\phi \in L^1\}$

take $\delta= \varepsilon/ \sum^n a_i$ where $\phi=\sum ^n a_iX_{A_i}$

now If $m(E)<\delta$ $\Rightarrow \int_R\phi X_E \:dμ=\sum ^n a_im({A_i\cap E})<\sum ^n a_i\delta=\varepsilon $

thus we get $ \int_E\phi \:dμ<\varepsilon $ and because $\int_Ef dμ=\sup \{\int_E\phi \:dμ\} \Rightarrow \int_Ef dμ< \varepsilon $

The proof is a little technical, so I have my doubts. Could someone verify that the proof is ok ?

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  • I even don't understand it... what is this $\sum^{n}a_i$ and this $\phi$ ?.... – Surb Aug 24 '22 at 13:37
  • @Surb $\phi $ is a simple function , I explain in the first row $\sum^n_{i=1} a_i$ are the values of $\phi$ – 領域展開 Aug 24 '22 at 13:40
  • this is just not a proof... – Surb Aug 24 '22 at 13:42
  • @Surb it's almost the same as the answer here https://math.stackexchange.com/questions/513401/for-every-epsilon0-there-exists-delta0-such-that-int-afx-mudx I don't understand what you mean, so I guess it's ok. – 領域展開 Aug 24 '22 at 13:44
  • If you don't see the difference between your proof, and the proof in the link that you posted, it's I think a big problem. Writing a mathematical proof don't consist in writing symbol randomly without any meaning a priori... But if for you it's ok, then... – Surb Aug 24 '22 at 13:57

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