Lebesgue measure on $L^2$

Question

This came from an old qualification exam for measure theory:

Suppose that $f:\mathbb{R}\to \mathbb{R}$ is monotonically increasing and absolutely continuous, and let $f'$ denote its Lebesgue a.e. define derivative.

a.) Show that if $f'\in L^2(m)$ ($m$ is Lebesgue measure on $\mathbb{R}$) then there is a $C < \infty$ such that for every Lebesgue measurable set $E$ $$m(f(E)) \leq C(m(E))^{1/2}$$

b.) Provide a counterexample to show that if we instead assume $f'\in L^1(m)$ then such a $C$ may not exist.

I normally don't post a question without some what of an attempt, but I really have no idea how to do this. Any suggestions or comments are greatly appreciated, I will edit this once I think of something.

Answer 1

In case of $(a)$, $$ m(f(E))=\int_{y\in f(E)}1\ dy. $$ Substitute $y=f(x)$. Then, $$ m(f(E))=\int_{x\in E}f'(x)\ dx\leq\left(m(E)\right)^{\frac{1}{2}}\left(\int|f'(x)|^{2}\ dx\right)^{\frac{1}{2}}=C\left(m(E)\right)^{\frac{1}{2}}. $$ Here, Holder inequality is used and $C=\left(\int|f'(x)|^{2}\ dx\right)^{\frac{1}{2}}<\infty$ since $f'(x)\in L^{2}$.

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For $(b)$, Hint : $$ f(x)=\frac{1}{\sqrt{x}} $$ with $E=(0,1)$.

Answer 2

The first one follows by Cauchy Schwarz: $(\int_Ef'dm)^2=(\int f'1_Edm)^2\leq (\int f'^2dm)(\int1_E^2dm)=||f'||_2^2(m(E))$ so take $C=||f'||_2$

For the next part, look at $f(x)=\dfrac{1}{\sqrt{x}}$ for $x\in (0,1]$ and $0$ otherwise. Then $f\in L^1(m)$ but $\notin L^2(m)$. Then $f((0,1])=[1,\infty)$ and clearly $m(f((0,1])=\infty$ implying there is no such $C$ as $m((0,1])=1$