Suppose $f$ is uniformly continuous on $[a,b]$ and $E\subset [a,b]$ such that $\mu(E)=0$ (lebesgue measure) then prove that $f(E)$ is of measure zero.
What i have tried so far is :
As $f$ is uniformly continuous, given $\epsilon>0$ there exists $\delta>0$ such that for $x,y\in [a,b]$ with $|x-y|<\delta$ we have $|f(x)-f(y)|<\epsilon$.
Let $E$ be a subset of measure $0$ then there exists an open set $E\subset U\subset [a,b]$ such that $\mu(U)<\delta$..
I want to show that $f(E)\subset f(U)$ and that $\mu(f(U))<\epsilon$ concluding that $f(E)$ is of measure zero.
Now, there exists $c,d\in U$ such that $f(c)\leq f(x)$ and $f(d)\geq f(x)$ for all $x\in U$.. ***
So, we have $\mu(f(U))=\left|f(d)-f(c)\right|<\epsilon$ as $c,d\in U$ and $\mu(U)<\delta$..
So, $f(E)\subset f(U)$ and $\mu(f(U))<\epsilon$..
So, given any $\epsilon>0$ we have $f(E)<\epsilon$. Thus $\mu(f(E))=0$..
Please let me know if there are any gaps..
Thanks..
*** such $c$ and $d$ exists when $U$ is a closed set.. But i thought i can consider closure of $U$ and choose $c,d$ from closure of $U$.. I believe $\mu(U)=\mu(\overline{U})$