# Approximation of measurable function via simple functions [Proof]

Let's denote by $$\mathfrak{M}$$ the set of all measurable subsets of $$\mathbb{R}^d$$. I mean the Lebesgue measure.

Definition: Let $$\{E_k\}_{k=1}^{N}\in\mathfrak{M}$$ with $$m(E_k)<\infty$$ then the function of the form $$\varphi(x)=\sum \limits_{k=1}^{N}a_k\chi_{E_K}(x)$$ is called simple function.

Theorem: Let $$f:X\to [0,+\infty]$$ is measurable function, then there exist the sequence of real-valued simple functions $$\{s_n(x)\}_{n=1}^{\infty}$$ on $$X$$ such that $$0\leq s_1\leq s_2\leq \dots\leq f$$ and $$s_n\to f$$ pointwise.

This is the theorem from Stein Shakarchi's book but I have found the following proof (not from the book).

Honestly to say two moments of the proof are quite not precise.

1) Note that we can write the function $$\phi_n(x)$$ in the following form $$\phi_n(x)=\sum \limits_{k=0}^{n2^n-1}\frac{k}{2^n}\chi_{\left[\frac{k}{2^n},\frac{k+1}{2^n}\right)}(x)+n\chi_{[n,+\infty)}(x)$$ But this function is not simple simple since the interval $$[n,+\infty)$$ has infinite measure.

2) When we compose any function on the left with simple function, namely $$\phi_n\circ f(x)$$ we get the following function $$\phi_n\circ f(x)=\sum \limits_{k=0}^{n2^n-1}\frac{k}{2^n}\chi_{f^{-1}\left[\frac{k}{2^n},\frac{k+1}{2^n}\right)}(x)+n\chi_{f^{-1}[n,+\infty)}(x),$$ since $$f$$ is measurable then we know that each set $$f^{-1}\left[\frac{k}{2^n},\frac{k+1}{2^n}\right)$$ and $$f^{-1}[n,+\infty)$$ is measurable but why their measure is finite?

Different authors use different definitions for simple functions, for example, Rudin defines $$s$$ to be simple if its range is finite. So, $$1_{[n,\infty)}$$ is simple in Rudin's world.
If you use the finite measure (and finite range) definition then try $$s'_n(x) = \phi_n(f(x)) \cdot 1_{[-n,n]^d}(x)$$. Then $$s_n'$$ has the same properties as $$s_n$$ except that you lose uniform convergence on sets on which $$f$$ is bounded.
• I accept your answer as the best! Really neat remark and I have checked it in the paper ;) But one moment: why it lose uniform convergence on sets where $f$ is bounded? – ZFR Oct 5 '18 at 23:51
• Note that in this excerpt we have the following result: $|s_n(x)-f(x)|<2^{-n}$ whenever $f(x)\in [0,\infty)$, right? Let $K=\{x: f(x)\leq M\}$ the set of points where $f(x)$ is bounded then we see that since this inequality does not depend on $x$ then we have uniform convergence on $K$, right? But the same reasoning can be applied even on the set of points where $f$ is finite namely $F=\{x: f(x)<\infty\}$. Am I true? – ZFR Oct 6 '18 at 0:04
• Take $f=1$ on $\mathbb{R}$, then $s_n'(x) = 1_{[-n,n]}(x)$ and hence $|f(n+1)-s_n'(n+1)| = 1$ for all $n$. Hence the convergence is not uniform unless $f$ has bounded support. Rudin's definition works because the sets involved in the simple function can have infinite measure. – copper.hat Oct 6 '18 at 1:59
• I understood your example with $f=1$ and we see that in this case we have not uniform convergence on the set where $f$ is bounded, in this case it's whole $\mathbb{R}$. But there are some questions: 1) I know that Rudin's definition of simple functions can have infinite measure. But it's a dumb question: but it's not obvious to me why in this case the convergence is uniform? Could you clarify this a bit? – ZFR Oct 6 '18 at 15:36