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Q1: How to construct a sequence of $\{f_n\}$ of simple function for a function $f$ such that $f_n\to f$ converges pointwise?

Q2: If $f$ is measurable is $f_n$ also measurable for each $n$?

Q2 is by Q1

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What does "p.c." mean? –  Davide Giraudo Jan 24 '13 at 12:42
    
pointwise coverge –  user54144 Jan 24 '13 at 17:57
    
See Theorem 2.10 in Folland's book "Real Analysis, Modern techniques and their applications". –  Vahid Shirbisheh Jan 24 '13 at 18:28

1 Answer 1

You can do it in a very direct way. Namely, for each $n$ choose a number of values allowable for the function $f_n$ - for instance the values of the form $\frac{k}{2^n}$ where $k$ is and integer and $-n 2^n \leq k \leq n 2^n$ (so that the value is $(-n,n)$). Now, at each point, assign the "best" possible value, by declaring for instance that $f_n(x) = \frac{k}{2^n}$ if $x \in \left[\frac{k}{2^n}, \frac{k+1}{2^n}\right)$, and $f_n(x) = \pm n$ if $\lvert f(x)\rvert > n$.

To see pointwise convergence, just note that if you fix $x$ and take $n > \lvert f(x)\rvert$ at some point $x$, then $\lvert f(x) - f_n(x) \rvert \leq \frac{1}{2^n}$. And $f_n$ are simple, because they take a finite number of values by definition.

If $f$ is measurable, then $f_n$ are measurable, because you can express preimages of the possible values of $f_n$ using preimages of measurable sets (intervals, even) of $f$.

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