Objective question on measure theory . Which of the following statements is/are true ?
$1.$ continuous image of a Lebesgue measurable set of $\Bbb R$ is Lebesgue measurable i.e. if $ f$ is continuous function and $A$ is Lebesgue measurable then $f(A)$ is also measurable.
$2.$ Lebesgue outer measure is invariant under continuous image i.e. if $f$ is a continuous function on real line then $m^*(A)=m^*(f(A))$.
$3.$ if $f$ is a homeomorphishm from $\Bbb R$ to $\Bbb R$ then $m^*(A)=m^*(f(A))$ for any set $A\subseteq \Bbb R$.
$4.$ if $f$ is a homeomorphishm from $\Bbb R$ to $\Bbb R$ then $m^*(A)<\infty\implies m^*(f(A))<\infty$ for any set $A\subseteq \Bbb R$.
I solved first three option as $1$ is false by Cantor function , second option is also false as constant function maps every set to measure zero set . For third option I choose  function $f(x)=3x$ for  counter example . I am unable to solve last option . Please help me in this option . Thanks.
 A: The last one is not true as well. Take $f$ s.t. $f(x)=e^x$ when $x>0$ (you can prolonge it as an homeomorphism over $\mathbb R$). Take $$A=\bigcup_{n=2}^\infty \underbrace{\left(n,n+\frac{1}{n^2}\right)}_{=:A_n}.$$
Then $m^*(A)<\infty $ but $m^*f(A))=\infty $ because $m^*f(A_n)=e^n\left(e^{\frac{1}{n^2}}-1\right)$, and $$\sum_{n\geq 2}e^n\left(e^{\frac{1}{n^2}}-1\right)=\infty .$$
A: Consider a set $$A=\bigcup_{n=1}^{\infty} \left(\sqrt{n} ,\sqrt{n+\frac{1}{n}}\right).$$
Since $$-\sqrt{n} +\sqrt{n+\frac{1}{n}}=\frac{\frac{1}{n}}{\sqrt{n} +\sqrt{n+\frac{1}{n}}} \leq \frac{1}{2(\sqrt{n})^3}$$ the set $A$ has finite measure an is obviously Lebesgue measurable.
Now consider a function $h:\mathbb{R}\to\mathbb{R}$ defined belov:
$$h(x) =\begin{cases} x^2 \mbox{ for } x\geq 0 \\ -x^2 \mbox{ for } x<0\end{cases}.$$
The function is an homeomorphism and $$h(A)=\bigcup_{n=1}^{\infty} \left(n ,n+\frac{1}{n}\right).$$
Hence the set is Lebesgue measurable and $$m^* (A) =\sum_{n-1}^{\infty } \frac{1}{n} =\infty .$$
Thus 4 is not true.
