Dear friends, let us use the definition of Lebesgue integral on ##X,\mu(X)<\infty## as the limit(adsbygoogle = window.adsbygoogle || []).push({});

##\int_X fd\mu:=\lim_{n\to\infty}\int_Xf_nd\mu=\lim_{n\to\infty}\sum_{k=1}^\infty y_{n,k}\mu(A_{n,k})##where ##\{f_n\}## is a sequence of simple, i.e. taking countably many values ##y_{n,k}## for ##k=1,2,\ldots##, functions ##f_n:X\to\mathbb{C}## uniformly converging to ##f##, and ##\{y_{n,k}\}=f_n(A_{n,k})## where ##\forall i\ne j\quad A_{n,i}\cap A_{n,j}=\emptyset##.

I know that if any sequence ##\{f_n\}\subset L_p(X,\mu)##, ##p\geq 1## uniformly converges to ##f## then [it also converges][1] with respect to the norm ##\|\cdot\|_p## to the same limit, which is therefore an element of ##L_p(X,\mu)##.

I read in Kolmogorov-Fomin'sElements of the Theory of Functions and Functional Analysis(1963 Graylock edition, p.85) thatan arbitrary function##f\in L_2##can be approximated[with respect to norm ##\|\cdot\|_2##, I suspect]with arbitrary simple functions belonging to##L_2##.

I do not understand how the last statement is deduced. If it were true, I would find the general case for ##L_p##, ##p\geq 1##, even more interesting. I understand that if ##f\in L_p\subset L_1## then for all ##\varepsilon>0## there exists a simple function ##f_n\in L_1## such that ##\forall x\in A\quad |f(x)-f_n(x)|<\varepsilon## and then, for all ##p\geq 1##, ##\|f-f_n\|_p<\varepsilon##, but how to find ##f_n\in L_p## (if ##p=2## or in general ##p>1##)?

Can anybody explain such a statement? Thank you so much!!!

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# Approximation of ##f\in L_p## with simple function ##f_n\in L_p##

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