Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I've come across this problem in my notes:

$\forall \sigma, \delta \geq 0:\ \exists \varphi$ continuous s.t. $\mu \{x: f(x)-\varphi(x)\geq \sigma\} \leq \delta$

Find $N: \frac{2M}{N} \leq \sigma$

let $E_i=\{x: \frac{(i-1)M}{N} \leq f(x) \leq \frac{iM}{N}\}\ i=1-N,..., N$

for each $E_i$ find $F_i \subset E_i:\ \mu F_i \leq \mu E_i- \frac{\delta}{4n}$

$F=\bigcup_{i=1}^N F_i$ where $\{F_i\}_{i=1-N}^N$ are disjoint closed sets

$\mu([a,b] \backslash F)\leq \frac{\delta}{2}$

we define $\varphi|_{F_i}=\frac{iM}{N},$ I understand that $\varphi$ is continuous on $F$. But how can we argue it is continuous on the whole interval $[a,b]$. What do we do about the endpoints?

share|cite|improve this question
up vote 1 down vote accepted

we need to show it's continuous at intervals endpoints, then: $\varphi|_{F_i+1}-\varphi|_{F_i} =\frac{M}{N}<\frac{2M}{N} \leq \sigma $

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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