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Approximate any bounded monotone function $f$ on a closed interval by a sequence of continuous functions that converge to $f$ almost everywhere?

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I am reading Real Variable and Integration: with historical note by Benedetto, and he said that any monotone decreasing and bounded real-valued function $f$ on a closed interval [a,b] can be approximated by a sequence of continuous monotone decreasing functions that converge to $f$$m$-a.e.

This is what I have so far:

Since $f$ is a monotone decreasing function, $f$ is measurable. Note also on $[a,b]$, the Lebesgue measure $m$ is regular, therefore, we can show as a corollary of Lusin's theorem that any measurable function $f$ which is finite a.e can be approximated by a sequence of continuous functions $\{f_{n}\}$ such that the convergence is m-a.e as follow:

By Lusin theorem, for each $n$, we can choose a continuous function $f_{n}$ such that $m(A_{n}) = m(\{x: f(x) \neq f_{n}(x)\}) < \frac{1}{2^n}$. By setting $A = \cap_{k=1}^{\infty}\cup_{n=k}^{\infty} A_{n}$, we have $m(A)=0$ since $m(\cup_{n=k}^{\infty}A_{n}) \leq \frac{1}{2^{k-1}}$. Also, by definition, $A$ is precisely the set of points $x$ which are in infinitely many $A_{k}$. Therefore, is $x \notin A$ then $x$ is in at most finitely many $A_{k}$ and so $f_{n}(x) = f(x)$ for all large $n$. Hence $f_{n} \to f$, m-a.e.

Now, define $g_{n}(x) = f_{n}(x) - \frac{x}{n}$. Then $g(x)$ is a continuous function as $f_{n}(x)$ and $\frac{x}{n}$ are continuous and are a decreasing sequence that converges to $f(x).


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