3
$\begingroup$

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?

$\endgroup$
5
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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