Proving that the sequence $φ_n:=\frac 1 n \sum_{k=0} ^{n^2} kχ_{E_{n,k}}$ is monotone

Question

I am trying to prove the following theorem (Royden 1st ed, Ch 11, Prop 10).

Let $(X,β,μ)$ be a $σ$-finite measure space and $f$ a nonnegative measurable function. Then there is a monotone increasing sequence $(φ_n)$ of simple functions, each vanishing outside a set of finite measure, such that $f=\lim φ_n$ at each point of $X$.

*I think that Royden intends that $\infty \notin img(f)$

I see that, being σ-finite, we can write $X=A_1 \cup A_2 \cup \cdots$, where each $A_i$ is measurable and has finite measure. Then we can define $X_i=A_1 \cup \cdots \cup A_i$. We see that $X_1\subseteq X_2 \subseteq \cdots$ and that each $X_i$ has finite measure and that $X=X_1 \cup X_2 \cup \cdots$.

Royden advises the reader to choose the sequence $φ_n:=\frac 1 n \sum_{k=0} ^{n^2} kχ_{E_{n,k}}$ where $E_{n,k}:= \{x \in X_n : \frac kn \leq f(x) < \frac {k+1}n \}$.

I have been able to prove that $\forall x\in X[\lim φ_n(x)=f(x)]$ and that $φ_n$ vanishes outside $X_n$ (a set of finite measure.) But I am having trouble proving that $φ_1\leq φ_2 \leq \cdots$.

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I have tried a proof by contradiction as follows. Assume there is a $n \in \mathbb{N}$ and a point $x\in X$ such that $φ_{n}(x)>φ_{n+1}(x)$. Since $φ_n(x)\neq 0$, $x\in E_{n,l}$ for exactly one $0\leq l \leq n^2$. We also see that $x\in E_{n+1,j}$ for exactly one $0\leq j \leq (n+1)^2$.

So we have the inequalities:

(1) $\frac l n \leq f(x) < \frac {l+1} n$

(2) $\frac j {n+1} \leq f(x) < \frac {j+1} {n+1}$

(3) $\frac l n> \frac j {n+1}$

I have tried manipulating these three inequalities to obtain a contradiction but, alas, I have yet to find one.

Answer 1

You can write $\phi_n(x)$ as $\frac {[nf(x)]} n$ (for $f(x)<n$) where $[a]$ is the greatest integer less than or equal to $a$. It is quite easy to find $a$ such that $\frac {[na]} n$ is not increasing, so the claim that the sequence $\{\phi_n\}$ is increasing is not correct. There is standard trick to get an increasing sequence $\phi_n$: consider the sets $\{\frac k {2^{n}} \leq f(x) <\frac {k+1} {2^{n}}\}$ ( $k \leq n2^{n}$) instead of the ones considered above. This makes the sequence increasing and this is what most text books on Measure Theory do.

In my copy of Royden (2nd edition) it does have $\{\frac k {2^{n}} \leq f(x) <\frac {k+1} {2^{n}}\}$ and not the sets you have quoted. Perhaps, a correction was made in later editions.

Answer 2

Ok,

So I believe the proof construction is incorrect - partially because Stein/Shakarchi uses a very similar construction in their if I recall (would need to confirm) and in my intro msr thry course we found that proof to be wrong.

Here's an example that convinces me: If $x$ is such that $\varphi_5$ is nonzero, and in fact is such that $\{ \frac{2}{5} \leq f(x) < \frac{3}{5} \}$ - meaning $k=2$, then $x$ is either such that $\{ \frac{2}{6} \leq f(x) < \frac{3}{6} \} = E_{2,6}$ or such that $\{ \frac{3}{6} \leq f(x) < \frac{4}{6} \} = E_{3,6}$. However, we are only monotone if $x \notin E_{2,6}$, for then $\varphi_{6}(x) = \frac{2}{6} < \frac{2}{5} = \varphi_5(x)$. However, we cannot guarantee for all $x \in E_{2,5}$ that $x\notin E_{2,6}$ without futher assumptions on $f$.

Agree?