I know that the question has been asked before, but I'm asking if I'm on the right track with my work, and also some further guidance to what to do next.
My work
First, let's suppose $f: \mathbb{R} \rightarrow [0, \infty)$. That is, assume that $f$ is non-negative.
We know that we can choose a sequence of non-negative simple functions such that
–$s_n \rightarrow f$ pointwise
–$(s_n(x))_{n=1}^\infty$ is increasing for all $x \in \mathbb{R}$
Claim:
$$f(x) = \sum_{n=1}^\infty (s_{n+1} – s_n)(x) + s_1(x)$$
Note that
$$\sum_{n=1}^\infty (s_{n+1} – s_n)(x) = \lim_{n \to \infty} \sum^n_{i = 1} (s_{i+1}- s_{i}) = \lim_{n \to \infty} s_{n+1}(x) – s_1(x) = f(x) – s_1(x)$$, from which the claim follows.
Hence, $f$ is a countable sum of non-negative simple function, so it may be written as
$$f = \sum_{n = 1}^\infty a_n \chi_{A_n}$$
where $a_n \in [0, \infty)$ and $A_n$ is Lebesgue measurable, and $\chi$ denote the characteristic function.
Now, since $A_n$ is measurable, we may choose $F_\sigma$(and hence Borel measurable) set $D_n$ such that $D_n \subseteq A_n$ and $m(A_n \setminus D_n) = 0$.
Let
$$h = \sum_{n = 1}^\infty a_n \chi_{D_n}$$
$h$ is Borel measurable(this should be right?) so it suffices to show the below claim.
Claim: $h = f$ almost everywhere.
And I'm stuck here, as the claim seems intuitively right, but I do not see a way to proceed. It would be simple if each $A_n$ was disjoint, but of course we can't guarantee that.
Any help would be appreciated, thanks!
Best Answer
First, note that we can find Borel measurable set $D_n$ and $C_n$ such that $D_n \subseteq A_n \subseteq C_n$ and $m(C_n \setminus D_n) = 0$. Define $h$ as follows, and show that if $x \in (\cup_{i=1}^\infty A_n \setminus D_n)^c$, then $f(x) = h(x)$.