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Assume $U(I)$ is all the functions $f:[a,b]\to \mathbb{R}$ such that there exists an increasing sequence $(\phi_n)_n$ of simple functions such that $ \lim\int_{a}^{b}\phi_n<\infty $ and $\phi_n\to f$ (almost everywhere). And assume $U_0(I)$ is all the functions $f:[a,b]\to \mathbb{R}$ such that there exists an increasing sequence $(\phi_n)_n$ of simple functions such that $ \lim\int_{a}^{b}\phi_n<\infty $ and $\phi_n\to f$.

Is it true that every Upper function(= every element of U(I)) is almost everywhere equal to an element of $U_0(I)$ ?

Darman
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  • by definition a simple function is finite everywhere and have finite integral, so the condition $\int \phi_n<\infty $ is redundant – Masacroso Nov 07 '20 at 11:00
  • @Masacroso Yes I can eliminate the condition $\exists \lim \int_{a}^{b}\phi_n$. But can it help me? – Darman Nov 07 '20 at 12:20
  • So the only difference between $U(I)$ and $U_0(I)$ is that in the latter, $\phi_n \to f$ everywhere, where as in the former $\phi_n \to f$ almost everywhere? – Paul Sinclair Nov 07 '20 at 16:20
  • @PaulSinclair Yes, exactly. (In fact, $\lbrace f-g:\ \ f,g\in U(I)\rbrace =$ all the Lebesgue integrable functions) – Darman Nov 07 '20 at 17:46

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If $\phi_n\to f$ almost everywhere this mean that $\mathbf{1}_{A}\phi_n\to \mathbf{1}_{A}f$ everywhere for some measurable set $A\subset \mathbb{R}$ such that $\mu(A^\complement )=0$ (here $\mu$ is the measure defined in $[a,b]$, it can be the Lebesgue measure or any other). Now note that $\mathbf{1}_{A}f=f$ almost everywhere.

Masacroso
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