# Let $A$ be a lebesgue null set in $\mathbb{R}$. Prove that $B:=\{e^x:x \in A\}$ is also a null set

Attempt:

Let $l$ be the Lebesgue measure. Since $A$ is a null set, there exists a countable open cover of intervals, $\{I_n\}_{n=1}^{\infty}$ such that $\sum_{n=1}^{\infty} l(I_n)<\epsilon$. Since $e^x$ is strictly increasing on $\mathbb{R}$. an interval $(a_n,b_n)$ is mapped to an interval $e^{I_n}=(e^{a_n},e^{b_n})$. I need to show that $\sum_{n=1}^{\infty} l(e^{I_n})<\epsilon$

Question: How should I proceed?

Well, it's definitely not true that $$\sum_{n = 1}^{\infty} \ell(e^{I_n}) < \epsilon$$ Consider an interval really far to the right, and notice that the derivative of $e^{x}$ blows up rather quickly. What you can say is that

$$\ell(e^{I_n}) \le e^{b_n} \ell(I_n)$$

You need some quantitative control on the exponential, which you currently do not have.

To fix this, we need to overcome the obstacle of "what if $b_n$ is really large"? One way to do this is to break up the sum into components where we do have good control. Using the above remark, you can show that

$$\{e^a : a \in A \cap [n, n + 1]\}$$

is a null set for each $n$: Choose a cover of $A \cap [n, n + 1]$ whose lengths sum to something like $e^{-(n + 1)} \epsilon 2^{-n}$, and the corresponding cover on the other side has lengths summing to at most $\epsilon 2^{-n}$.

Now either sum over $n$, or just notice that a countable union of null sets is null.