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?

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1 Answer 1


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.

  • $\begingroup$ 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? $\endgroup$
    – RFZ
    Commented Oct 5, 2018 at 23:51
  • $\begingroup$ 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? $\endgroup$
    – RFZ
    Commented Oct 6, 2018 at 0:04
  • 1
    $\begingroup$ 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. $\endgroup$
    – copper.hat
    Commented Oct 6, 2018 at 1:59
  • $\begingroup$ 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? $\endgroup$
    – RFZ
    Commented Oct 6, 2018 at 15:36
  • 1
    $\begingroup$ I understood it completely! Thanks a lot for that! :) $\endgroup$
    – RFZ
    Commented Oct 6, 2018 at 17:09

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