13
$\begingroup$

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

Let $f:E \rightarrow \mathbb{R}$ be measurable. There exist simple measurable functions $\phi_n$ on $E$ such that $$(a) \ \phi_1 \leq \phi_2 \leq \cdots \leq f.$$ $$(b) \ \forall x\in E, \text{ we have that } \phi_n(x) \rightarrow f(x), \text{ as }n \rightarrow \infty$$ Proof:

Notice that for each positive integer $n$ and each real number $t$ corresponds a unique integer $k = k_n(t)$ that satisfies $k2^{-n} \leq t < (k+1)2^{-n}$. Define: $$ s_n(t) = \left\{\begin{matrix} k_n(t)2^{-n} & 0 \leq t < n\\ n & n \leq t \leq \infty \end{matrix}\right. $$ Note that each $s_n$ is a borel function on $[0,\infty]$. In other words, for any open set $V$, $s_n^{-1}(V)$ is a borel set. Now, $$k2^{-n} \leq t < (k+1)2^{-n} \Rightarrow t - 2^{-n} < s_n(t) \leq t \text{ if } 0 \leq t\leq n,$$ thus, $0 \leq s_1 \leq s_2 \leq \cdots \leq t,$ and $s_n(t) \rightarrow t$ as $n \rightarrow \infty,$ for every $t\in [0,\infty]$. It follows that the function $\phi_n = s_n\circ f$ satisfy (a) and (b); since $f$ is measurable and $s_n$ is a borel function, then $\phi_n$ is also measurable. To obtain a decreasing function, let $\psi_n = -s_n(-f)$, thus $\phi_n$ and $\psi_n$ are steps functions, $ \phi_n \leq f\leq \psi_n$ and $\psi_n-\phi_n \leq 2^{-n}$ for every integer $n$.

My question is how I can obtain uniform convergence? I know that from what i proved, I didn't needed to assume that $f$ is bounded.

The following theorem might be useful, namely Dini's Theorem(obtained from wikipedia).

If $Y$ is a compact topological space, and $\{f_n\}_{n\in \mathbb{N}}$ is a monotonically increasing sequence (meaning $f_n(x) \leq f_{n+1}(x)$ for all $n$ and $x$) of continuous real-valued functions on $X$ which converges pointwise to a continuous function $f$, then the convergence is uniform. The same conclusion holds if $\{f_n\}_{n\in \mathbb{N}}$ is monotonically decreasing instead of increasing.

Is there a way to use this theorem under the assumption that $f$ is bounded to obtain uniform convergence?

$\endgroup$
4
  • $\begingroup$ $f$ bounded does not imply $E$ (or $X$, as your post uses both letters for the domain of $f$) is compact. Are you sure that this wasn't assumed somewhere earlier? $\endgroup$ Commented Oct 29, 2015 at 1:07
  • 2
    $\begingroup$ This is a standard result that appears in zillions of real analysis texts, and has nothing to do with topology. $\endgroup$
    – zhw.
    Commented Oct 29, 2015 at 2:01
  • $\begingroup$ I updated the problem description, hope it helps. $\endgroup$
    – okie
    Commented Oct 29, 2015 at 3:36
  • $\begingroup$ After the question the theorem you have written starting as "Let...... " , in that theorem you have $\phi_{1}, \phi_{2}, ... $ a sequence of increasing functions that converges to $f$ pointwise, but that is possible when f is non-negative. How do you get that theorem for arbitrary $f$. $\endgroup$
    – Riju
    Commented Jul 19, 2017 at 2:11

1 Answer 1

9
$\begingroup$

Choose $N \in \mathbb{N}$ sufficiently large such that $\|f\|_{\infty}<N$. Then it follows from the very definition of $\phi_n$ that

$$|\phi_n(x)-f(x)| \leq 2^{-n}$$

for all $x$ and $n \geq N$. This means that $\|\phi_n-f\|_{\infty} \to 0$ as $n \to \infty$.

$\endgroup$
4
  • $\begingroup$ Is this enough to prove uniform convergence? A quick question, $\|f\|_{\infty}<N$ means that $f$ is bounded, right? $\endgroup$
    – okie
    Commented Oct 29, 2015 at 19:59
  • $\begingroup$ @RicardoCervantes It means that $|f(x)|<N$ for all $x$ (in particular, $f$ is bounded). Concerning your first question: What does uniform convergence of a sequence mean? What do you have to show? $\endgroup$
    – saz
    Commented Oct 29, 2015 at 21:00
  • $\begingroup$ i need to prove, given $\epsilon > 0 \exists N$, such that $|\phi_n(x) - f(x)| < \epsilon$ for every $n \geq N$ and for all $x \in E$. $\endgroup$
    – okie
    Commented Oct 29, 2015 at 22:38
  • $\begingroup$ @RicardoCervantes Exactly. So, for fixed $\epsilon>0$,we choose $N$ sufficiently large such that $\|f\|_{\infty}<N$ and $2^{-N}<\epsilon$. Then, we have $$|\phi_n(x)-f(x)| \leq 2^{-n}<\epsilon$$ for all $x$ and $n \geq N$. $\endgroup$
    – saz
    Commented Oct 30, 2015 at 6:08

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .