4
$\begingroup$

Show that lower semicontinuous function $f:X\rightarrow [0,1]$ on metrizable X is the supremum of an increasing sequence of continuous functions.

My attempt: I don't know how to approximate $f(x)$ to within $[f(x)-1/n,f(x)]$ by $h_n(x)$ which is a linear combination of characteristic function of open sets, using lower semicontinuity of $f$.

Can anyone help?

$\endgroup$
1
  • 1
    $\begingroup$ Now sure how to proceed with your idea. But from a post a couple years back... Let $f_k(x)=\inf\{f(y)+kd(x,y):y\in X\}.$ $\endgroup$ – matt biesecker May 13 '15 at 5:00
2
$\begingroup$

This is a quotation from "General Topology" by Ryszard Engelking:

enter image description here

$\endgroup$
0
$\begingroup$

This is a problem from Willard's general topology textbook. I shall provide an outline of the proof and leave the details to the reader.
The function $f(x)$ may be approximated to within $[f(x)-1/n, f(x)]$ by the function $$h_{n} = \frac{1}{n}\sum_{k = 1}^{n-1}\chi_{f^{-1}\left(\left(\frac{k}{n}, 1\right]\right)}$$ where $\chi_{A}$ denotes the characteristic function of a subset $A$ of $X$. Note that $h_{n}$ is a linear combination of characteristic functions of open sets.
The characteristic function of any open set $A$ in a metric space $(X, \rho)$ can be written as the supremum of an increasing sequence of continuous functions. To prove this, note that every open set in a metric space is $F_{\sigma}$. In particular, if $A$ is an open set then $A$ can be written as the union of an increasing sequence $A_{1}\subseteq A_{2}\subseteq\cdots$ of closed sets where $$A_{n} = \left\{y\in X: \rho(y, X-A)\geq \frac{1}{n}\right\}.$$ For each positive integer $n$ define $g_{n}:X\rightarrow\mathbb{R}$ by $g_{n}(x) = \min\{n\rho(x, X-A), 1\}$ and verify that $\chi_{A} = \sup_{n}g_{n}$.
For each $1\leq k\leq n-1$ let $$\frac{1}{n}\chi_{f^{-1}\left(\left(\frac{k}{n}, 1\right]\right)} = \sup_{m}g_{k,m}$$ for an increasing sequence of continuous functions $\{g_{k, m}\}_{m\in\mathbb{Z}^{+}}$. One can verify that $h_{n} = \sup_{m} h_{n, m}$ where $h_{n, m} = \sum_{k = 1}^{n-1}g_{k,m}$.
Finally, one can show that $$f = \sup_{(n, m)\in\mathbb{Z}^{+}\times\mathbb{Z}^{+}}h_{n, m}.$$ Using any bijection between $\mathbb{Z}^{+}\times\mathbb{Z}^{+}$ and $\mathbb{Z}^{+}$, rearrange the functions $\{h_{n,m}\}$ into a sequence $\{\tilde{k}_{n}\}$ and let $k_{n} = \max\{\tilde{k}_{1},\dots,\tilde{k}_{n}\}$. Then $\{k_{n}\}$ is the required sequence of increasing continuous functions such that $f = \sup_{n}k_{n}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.