# Question related to approximation of measurable functions by simple functions

I am self-studying measure theory from the book by Sheldon Axler.There I found a theorem before integration is introduced:

Let $$(X,\mathcal S)$$ be a measurable space and $$f:X\to [-\infty,\infty]$$ be a measurable function.Then there exists a sequence of simple measurable functions $$(f_n)$$ such that:

$$1.$$ $$|f_n(x)|\leq |f_{n+1}(x)|\leq |f(x)|$$ for all $$n\in \mathbb N$$ and $$x\in X$$.

$$2.$$ $$f_n\to f$$ pointwise on $$X$$.

Now I have a number of questions regarding this theorem:

First one is that why the codomain of $$f$$ is taken as $$[-\infty,\infty]$$ instead of $$\mathbb R$$.Will it affect the theorems related to integrals,if we choose $$\mathbb R$$?

Second one is that why is modulus sign given in $$(1)$$.Can I simply omit the moduli by taking $$f\geq 0$$.

Now comes the next section of the question,about the proof of this theorem(even if it is assumed that codomain is $$\mathbb R$$ and $$f\geq 0$$).I still do not get how to prove this result.Can someone guide me a bit by providing intuition so that I can visualize the idea behind the proof,because the proof I found in the book is a bit technical and constructive.

• There is some repetition in 1., perhaps one of the indices is off by 1? Apr 7 at 16:22

The intuition is that you are essentially discretizing $$f$$ with granularity $$h > 0$$: Let $$f : X \to [0, \infty]$$ be measurable. Let $$h > 0$$. Define $$f_h : X \to [0, \infty]$$ by $$f_h(x) = \lfloor\frac{f(x)}{h}\rfloor h.$$ It's clear that $$f_h \nearrow f$$ as $$h \to 0$$. $$f_h$$ takes the values $$\{0, h, 2h, 3h, \dots\} \cup \{\infty\}$$. So $$f_h$$ is not a simple function since it takes on countably many values instead of just finitely many. To fix this, use $$\tilde{f_h} = f_h \land \alpha(h)$$, where $$\alpha(h)$$ is any function with $$\alpha(h) \nearrow \infty$$ as $$h \to 0$$, e.g. $$\alpha(h) = 1/h$$. Now $$\tilde{f_h}$$ takes the values $$\{nh : n \in \mathbb{N}, n \leq \alpha(h)\}$$, which is a finite set, and $$\tilde{f_h} \nearrow f \land \infty = f$$ as required.
The most common proof of the above approximation theorem uses the sequence $$f_n = \left(\lfloor\frac{f(x)}{2^{-n}}\rfloor 2^{-n}\right) \land n.$$ In my notation above, this corresponds to $$h = 2^{-n}$$ and $$\alpha(h) = -\log_2(h) = \log_2(1/h)$$.
• Please correct me if I am wrong.I am defining $f_k\to f$ as follows: Apr 9 at 3:28
• For $k\in \mathbb N$,define $f_k(x)=k$ if $f(x)>k$ and and if $0\leq f(x)<k$ then $f_k(x)$ is defined in the following manner: We first divide each interval of the form $[n,n+1)$ within $[0,k]$ to $2^k$ subintervals and define $f_k(x)$ to be the left endpoint of the subinterval in which $f(x)$ falls. Apr 9 at 3:32
• I want to know why $f_k$ is measurable provided $f$ is measurable. Apr 9 at 3:37
• @KishalaySarkar You have $f_k = \left(\lfloor\frac{f(x)}{2^{-k}}\rfloor 2^{-k}\right) \land k$. So you just have to show that the floor function is measurable and that the minimum of two measurable functions is measurable. Both of these are straightforward. Apr 9 at 4:16
• what about my first question?Can we replace the codomain by $\mathbb R$? Apr 9 at 8:11