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A quick question. Let $X \subset \mathbb R$ be a Lebesgue measurable set. Is it true that the set $2X = \{2 x : x \in X\}$ has measure $m(2X) = 2\ m(X)$? It is easy to prove it if for example if $X$ is an open set, but is it true in general? Thanks

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up vote 4 down vote accepted

Other users' proofs also work, but they're missing a more elegant method of proof, taking advantage of the one-to-one correspondence between covers of $X$ and covers of $2X$.

Lebesgue measure is defined as $m(X) = \inf\big\{\sum_n m(I_n)\big\}$ where the $I_n$ are intervals and $X$ is contained in their union.

For any such cover $\{I_n\}$ of $X$, $\{2I_n\}$ covers $2X$. The Lebesgue measure being a lower bound of the lengths of all such covers of $2X$, $$m(2X)\le\sum_n m(2I_n) = 2\sum_nm({I_n}).$$

Since this is true for all such $\{I_n\}$, $$m(2X)\le\inf\Big\{\sum_n m(I_n)\Big\}=2m(X).$$

And a similar argument proves the converse.

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I believe Brandon hints at exactly the same solution. – tomasz Jan 17 '14 at 7:34
This direction is correct but the converse inequality seems a bit more difficult – user61581 Jan 18 '14 at 2:17

Consider the $\sigma$-algebra generated by the open sets, $\mathcal{B}(\mathbb{R})$. Then consider $\lambda:\mathcal{B}(\mathbb{R})\to\mathbb{R^*}$ given by $\displaystyle\lambda(E):=\frac{m(2E)}{2}$. You can check that this is a measure on $\mathcal{B}(\mathbb{R})$. Moreover since you have shown that $m(2E)=2m(E)$ for open sets, you get that $\lambda(E)=m(E)$ on $\mathcal{B}(\mathbb{R})$. Lastly $\lambda^*$ is $\sigma$-finite, so by Carathéodory's extension theorem this extends to a unique measure on the $\sigma$-algebra of $\lambda^*$ measurable sets. But $m$ is also $\sigma$-finite and thus extends to the same measure. I.e. $\frac{m(2E)}{2}=m(E)$ on the Lebesgue measurable sets as required.

Replacing $2$ with $n\in\mathbb{R}_{\geq0}$ gives a slightly more general result.

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In general, if $A$ is linear and $X$ measurable, we have $m(AX) = |\det A| mX$.

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You can use that for any measurable set $X$, we have $$ m(X)=\inf\{\sum_{n=1}^\infty m(I_n): X\subset \bigcup_{n=1}^\infty I_n\} $$ Where the $I_n$ are understood to be open bounded intervals.

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Let $\chi_{E}$ be the characteristic function of a Lebesgue measurable set $E$ of finite Lebesgue measure. Then, using your notation $\chi_{2E}(2x)=\chi_{E}(x)$. $$ m(2E)=\int_{\mathbb{R}}\chi_{2E}(x)\,dx = 2\int_{\mathbb{R}}\chi_{2E}(2u)\,du=2\int_{\mathbb{R}}\chi_{E}(u)du=2m(E). $$ The case of infinite measure for either $E$ or $2E$ definitely implies the infinite measure of both; otherwise starting with the finite one, you could do the above change of variables and conclude that the other is of finite measure, too, with $m(2E)=2m(E)$.

It is important to remember that $L^{1}(\mathbb{R})$ is the completion of Riemann integrable functions under the integral norm, and that the Lebesgue integral is the unique continuous extension of the Riemann integral to $L^{1}(\mathbb{R})$. So, by design, all change of variables formulas for the Riemann integral continue to hold for the Lebesgue integral. People tend to forget how things are designed.

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