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 countable base 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 says 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... Thank you so much!!!


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