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?

enter image description here


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? $\endgroup$ – ZFR Oct 5 '18 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? $\endgroup$ – ZFR Oct 6 '18 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 Oct 6 '18 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? $\endgroup$ – ZFR Oct 6 '18 at 15:36
  • 1
    $\begingroup$ I understood it completely! Thanks a lot for that! :) $\endgroup$ – ZFR Oct 6 '18 at 17:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.