- #1
DavideGenoa
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Let ##f:X\to \mathbb{Q}+i\mathbb{Q}\subset\mathbb{C}##, ##f\in L_1(X,\mu)## be a Lebesgue-summable function taking only finitely many values ##y_1,\ldots,y_n\in \mathbb{Q}+i\mathbb{Q}## on the sets ##E_1,\ldots,E_n## such that ##\bigcup_{i=1}^nE_i=X##, ##\forall i\ne j\quad E_i\cap E_j=\emptyset##. Space ##X## is endowed with measure ##\mu##, which has a http://librarum.org/book/10022/196 that is a ring of sets.
By definition of countable base, for any ##E_k, k=1,\ldots,n## and any ##\varepsilon>0## there is a set of the basis ##A_k## such that ##\mu(E_k\triangle A_k)<\varepsilon##. Let us define ##A_k':=A_k\setminus\bigcup_{i<k} A_i## and the function
##f^\ast(x)= \begin{cases} y_k & \text{if } x\in A_k' \\ 0 & \text{if } x\in X\setminus\bigcup_{i=1}^n A_i' \end{cases}##
Then, Kolmogorov and Fomin's Introductory Real Analysis http://librarum.org/book/10022/196 that, "clearly", for ##\varepsilon## small enough, ##\mu\{x\in X:f(x)\ne f^\ast (x)\}## can be made arbitrarily small.
Could anybody show how to prove that ##\mu\{x\in X:f(x)\ne f^\ast (x)\}<\delta## for some choice of ##A_k##'s ?
I notice that, since ##(\bigcup_{k=1}^n E_k)\triangle(\bigcup_{k=1}^n A_k')=(\bigcup_{k=1}^n E_k)\triangle(\bigcup_{k=1}^n A_k)\subset\bigcup_{k=1}^n(E_k\triangle A_k)##, the inequality ##\mu((\bigcup_{k=1}^n E_k)\triangle(\bigcup_{k=1}^n A_k'))\leq\sum_{k=1}^n\mu(E_k\triangle A_k)<n\varepsilon ## holds. We also have that ##f^\ast(x)\ne f(x)\Rightarrow x\in X\setminus\bigcup_{k=1}^n(A_k'\cap E_k)=\bigcap_{k=1}^n (A'^c_k\cup E_k^c) ## where I use the notation ##S^c:=X\setminus S##. An inclusion of ##\bigcap_{k=1}^n (A'^c_k\cup E_k^c) \subset M## such that ##\mu(M)## is arbitrarily small would prove the desired result, but I am not sure we can find one...
Thanks such that ##\mu##(my thanks)##=+\infty##! ;)
By definition of countable base, for any ##E_k, k=1,\ldots,n## and any ##\varepsilon>0## there is a set of the basis ##A_k## such that ##\mu(E_k\triangle A_k)<\varepsilon##. Let us define ##A_k':=A_k\setminus\bigcup_{i<k} A_i## and the function
##f^\ast(x)= \begin{cases} y_k & \text{if } x\in A_k' \\ 0 & \text{if } x\in X\setminus\bigcup_{i=1}^n A_i' \end{cases}##
Then, Kolmogorov and Fomin's Introductory Real Analysis http://librarum.org/book/10022/196 that, "clearly", for ##\varepsilon## small enough, ##\mu\{x\in X:f(x)\ne f^\ast (x)\}## can be made arbitrarily small.
Could anybody show how to prove that ##\mu\{x\in X:f(x)\ne f^\ast (x)\}<\delta## for some choice of ##A_k##'s ?
I notice that, since ##(\bigcup_{k=1}^n E_k)\triangle(\bigcup_{k=1}^n A_k')=(\bigcup_{k=1}^n E_k)\triangle(\bigcup_{k=1}^n A_k)\subset\bigcup_{k=1}^n(E_k\triangle A_k)##, the inequality ##\mu((\bigcup_{k=1}^n E_k)\triangle(\bigcup_{k=1}^n A_k'))\leq\sum_{k=1}^n\mu(E_k\triangle A_k)<n\varepsilon ## holds. We also have that ##f^\ast(x)\ne f(x)\Rightarrow x\in X\setminus\bigcup_{k=1}^n(A_k'\cap E_k)=\bigcap_{k=1}^n (A'^c_k\cup E_k^c) ## where I use the notation ##S^c:=X\setminus S##. An inclusion of ##\bigcap_{k=1}^n (A'^c_k\cup E_k^c) \subset M## such that ##\mu(M)## is arbitrarily small would prove the desired result, but I am not sure we can find one...
Thanks such that ##\mu##(my thanks)##=+\infty##! ;)
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