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.

  • $\begingroup$ There is some repetition in 1., perhaps one of the indices is off by 1? $\endgroup$
    – copper.hat
    Apr 7 at 16:22

1 Answer 1


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 $\alpha(h) = -\log_2(h) = \log_2(1/h)$.

  • $\begingroup$ Please correct me if I am wrong.I am defining $f_k\to f$ as follows: $\endgroup$ Apr 9 at 3:28
  • $\begingroup$ 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. $\endgroup$ Apr 9 at 3:32
  • $\begingroup$ I want to know why $f_k$ is measurable provided $f$ is measurable. $\endgroup$ Apr 9 at 3:37
  • $\begingroup$ @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. $\endgroup$
    – Mason
    Apr 9 at 4:16
  • $\begingroup$ what about my first question?Can we replace the codomain by $\mathbb R$? $\endgroup$ Apr 9 at 8:11

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