Let $f$ be a bounded measurable function on $E$. Show that there are sequences of simple functions on $E$, $\{\varphi_{n}\}$ and $\{\psi_{n}\}$ such that $\{\varphi_{n}\}$ is increasing, $\{\psi_{n}\}$ is decreasing, and each of these sequences converge to $f$ uniformly on $E$.
In order to approach this problem, I was given the hint of using the Simple Approximation Lemma. This Lemma guarantees to us the existence of an increasing sequence $\{\gamma_{n}\}$ that converges pointwise to $f$ on $E$.
Then, and this is the first part of my question: Since $f$ is bounded, we have $|f(x)|\leq M$ $\forall x$. So, the function $f+M$ is both bounded and measurable, and we are given that $\{\gamma_{n}\}$ converges pointwise to $f$, and is increasing. Now, can we then let $\varphi_{n} = \gamma_{n} – M$, where $\{\varphi_{n} \}$ is a sequence that increases pointwise to $f$? And for another sequence $\{\eta_{n}\}$ increasing pointwise to $f$, can we let $\psi_{n} = -\eta_{n} $, and have $\{\psi_{n}\}$ be a sequence decreasing to $f$?
The second part of my question is: once I have my two sequences converging pointwise, how do I show that they converge uniformly? I believe it has something to do with the fact that $f$ is bounded $\forall x$ by a constant.
Best Answer
For the first part, if $\gamma_n\to f$, then $ (\gamma_n-M)\to (f-M)$. So unless $M=0$, in which case $f\equiv 0$, $\varphi_n=\gamma_n-M \not\to f$. Similarly $\psi_n\to -f$.
For the second part let me first recall the Simple Approximation Lemma you are citing:
The Simple Approximation Lemma: Let $f:E\subseteq\mathbb{R}\to\mathbb{R}$ be bounded and measurable. Then $\forall \epsilon>0,\exists$ simple $\varphi_\epsilon,\psi_\epsilon:E\to\mathbb{R}: \varphi_\epsilon\leq f\leq \psi_\epsilon$ and $ \psi_\epsilon-\varphi_\epsilon<\epsilon$ on $E$.
Observe that the result of the lemma already includes some uniformity, meaning that the difference of functions are bounded uniformly for $x\in E$. Thus once you construct the simple sequences and ensure that they are monotone you are done.
Also note that the boundedness of $f$ is required for the construction of simple functions in the lemma to begin with.
For reference purposes this is Exercise 3.2.12 from Royden & Fitzpatrick's Real Analysis, 4e (p. 63).