8

The following is from page 31 of Stein and Shakarchi's Real Analysis. My question is about an aspect of the proof of the following theorem.

Theorem 4.1 Suppose $f$ is a non-negative measurable function on $\mathbb R^d$. Then there exists an increasing sequence of non-negative simple functions $\{\varphi_k\}_{k=1}^\infty$ that converges pointwise to $f$, namely, $$ \varphi_k(x) \le \varphi_{k+1}(x)\quad\text{and}\quad\lim_{k\to\infty}\varphi_k(x)=f(x),\ \text{for all $x$.} $$ Proof. We begin first with a truncation. For $k\ge 1$, let $Q_k$ denote the cube centered at the origin and of side length $k$. Then we define $$ F_k(x) = \begin{cases} f(x) & \text{if $x\in Q_k$ and $f(x)\le k$,} \\ k & \text{if $x\in Q_k$ and $f(x)> k$,}\\ 0 & \text{otherwise.} \end{cases} $$ Then $F_k(x)\to f(x)$ as $k$ tends to infinity for all $x$. Now, we partition the range of $F_k$, namely $[0,k]$ as follows. For fixed $k,j\ge 1$, we define $$ E_{\ell,j}=\left\{x\in Q_k:\frac{\ell}{j}<F_k(x)\le\frac{\ell+1}{j}\right\},\quad\text{for}\ 0\le\ell<kj. $$ Then we may form $$ F_{k,j}(x) = \sum_{\ell=0}^{kj-1}\frac{\ell}{j}{\large{\chi_{E_{\ell,j}}}}(x) $$ [where $\large{\chi_{E_{\ell,j}}}$ is the indicator function of $E_{\ell,j}$].

Each $F_{k,j}$ is a simple function that satisfies $0\le F_k(x)-F_{k,j}(x)\le 1/j$ for all $x$. If we now choose $j=k$, and let $\varphi_k = F_{k,k}$, then we see that $0\le F_k(x)-\varphi_k(x)\le 1/k$ for all $x$, $\color{red}{\underline{\color{black}{\text{and $\{\varphi_k\}$ satisfies all the desired properties.}}}}$

I do not see why $\varphi_k(x)\le\varphi_{k+1}(x)$ for all $x$. Can someone explain that?

Alex Ortiz
  • 26,211

1 Answers1

6

Unless I have misunderstood something, I suspect there is a slight error in the proof. Consider $f(x)=x$ and consider the approximation of the point $x_0=.6$. Then for $k=2$ we have $x_0\in E_{1,2}$ and so $F_{2,2}(.6) = \frac{1}{2}$. However, for $k=3$ we have $x_0 \in E_{1,3}$ so that $F_{3,3}(.6)=\frac{1}{3}$. If we consider the sequence

$$\varphi_k(x) = F_{2^k,2^k}(x)$$

then I believe we have the desired result. Indeed, in this case we use the construction of $F_{k}$ to deduce that $F_{k-1}\leq F_{k}$ and the fact that the "vertical mesh" is finer for larger $j$ to deduce that $F_{2^k,2^{k-1}} \leq F_{2^k,2^k}.$ With these observations together we conclude

$$F_{2^{k-1},2^{k-1}}\leq F_{2^{k},2^{k-1}}\leq F_{2^k,2^k}.$$

Edit: to prove $F_{2^{k},2^{k-1}}\leq F_{2^k,2^k}$, fix $x$ and observe that if $x\in E_{\ell,2^{k-1}}$ then either $x\in E_{2\ell,2^{k}}$ or $x\in E_{2\ell+1,2^k}.$ We conclude

$$F_{2^{k},2^{k-1}}(x) = \frac{\ell}{2^{k-1}} \leq \min\{\frac{2\ell}{2^{k}},\frac{2\ell+1}{2^{k}}\}\leq F_{2^k,2^k}(x).$$

Matt
  • 2,267
  • I think you're right! Would you mind proving that first inequality $\large F_{2^k,2^{k-1}}\le F_{2^k,2^{k}}$? I feel like it should be true, but I am having a hard time putting it together formally. – Alex Ortiz Apr 18 '17 at 01:52
  • @AOrtiz I have updated my answer to prove this inequality. I have some pictures in my head about how these inequalities should work, but I am subject to human error; be sure to check me! – Matt Apr 18 '17 at 02:09
  • Yes! I just checked it. Your strategy dyadically partitions up the range, and the characteristic functions just keep getting larger because with each additional power of $2$, you refine the same subset of the range. Thanks very much. – Alex Ortiz Apr 18 '17 at 02:27
  • You're welcome! It was a good question... I also learned from Stein and Shakarchi, and I am sure I did not notice this error when learning this result. – Matt Apr 18 '17 at 02:29
  • Isn't there also an error in the definition of $F_{N, M}$? Surely for l=0 the function vanishes so there is no use – R. Reed Jan 06 '25 at 11:42