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Let $f$ be a bounded measurable function on $E$. Show that there are $\{\phi_n(x)\}$ and $\{\psi_n(x)\}$ - sequences of simple functions on $E$ such that $\{\phi_n(x)\}$ is increasing and $\{\psi_n(x)\}$ is decreasing and each of these converge to $f$ uniformly on $E$.

By the simple approximation lemma I know that both of the mentioned sequences exists such that $\phi_n(x) \leq f \leq \psi_n(x)$ and $\psi_n(x)-\phi_n(x) < \frac{1}{n}$ for each $n \in \mathbb{N}$.

Using uniform continuity: for some $\varepsilon > 0$ there exists an $n \geq N$ such that $|\phi_n(x)-f| < \frac{1}{n}$. Since simple functions take on a finite number of values, it is reasonable to take a $\max$. So I let $\phi(x)=\max\{\phi_n(x)\}$.

I guess I'm not sure how to pull together the lemma and the definition of simple function.

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up vote 3 down vote accepted

For every nonnegative integer $n$, define $A_n:\mathbb R\to\mathbb R$ by $A_n(t)=k2^{-n}$ for every $t$ such that $k\leqslant2^nt\lt k+1$, for some integer $k$ (in other words, $2^nA_n(t)$ is the integer part of $2^nt$).

Let $\phi_n=A_n(f)$ and $\psi_n=-A_n(-f)$. Then $\phi_n$ and $\psi_n$ are step functions, $\phi_n\leqslant f\leqslant \psi_n$, and $\psi_n-\phi_n\leqslant2^{-n}$ for every $n$. Furthermore, the sequence $(\phi_n)_n$ is nondecreasing and the sequence $(\psi_n)_n$ is nonincreasing.

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I take it your second paragraph is intended to get monotonic sequences out of $\phi$ and $\psi$. Probably if you look at your proof of the simple approximation lemma they already are monotonic.

If instead you want to define a new sequence $\phi'_n=\max_{i\leq n}\{\phi_n\}$ you can show $\phi'$ is simple by showing it takes only one value on each of the intersections of the sets on which the $\phi_i$ take different values. More precisely, let $\phi_i=\sum_{k=1}^{k(i)}a_{i,k}\chi_{E_{i,k}} $. Then the proposed $\phi'_n$ is constant on the sets $E_{1,k_1}\cap E_{2,k_2}\cap...\cap E_{n,k_n}$: since each of the $\phi_i$ are constant on such sets, their max can't change.

As you can see everything goes the same for defining $\psi'$ out of $\psi$.

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Should it be the unions of the $E_{i,k}$'s? My understanding is that we "chop up" the range of the function and look at the preimages of each interval. – emka Oct 18 '12 at 15:56
I don't understand any sense of understanding that requires quotation marks, but $\phi'_n$ definitely isn't constant on the unions. Take $\phi_1=\chi_{[0,1]}$ and $\phi_2=2\chi_{[-1,0)}+\chi_{[0,1]}$. Then our proposed $\phi'_2=\phi_2$, since in this example the sequence is monotonic. We have $E_{1,0}=\mathbb{R}\setminus [0,1],E_{1,1}=E_{2,2}=[0,1],E_{2,1}=[-1,0], E_{2,0}=\mathbb{R}\setminus [-1,1],E_{1,1}\cup E_{2,1}=[-1,1],$ on which $\phi'_2$ is not constant. – Kevin Carlson Oct 18 '12 at 16:07

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