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Can someone please explain: Assume for each $x \in \mathbb{R}$ and $A \subseteq \mathbb{R}$, that $x + A = \big\{ x + a \mid a \in A \big\}$. A and x + A are Borel sets for all $x \in \mathbb{R}$

Then, if λ is the Lebesgue measure on B, how can it be proven to be a translation invariant? So far all I have gotten is that λ(A) = λ(x + A), for all Borel sets A and for all $x \in \mathbb{R}$.

Eugene Zhang
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Kimmy634
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  • Isn't this the definition of a measure invariant by translation? – Tryss Mar 21 '16 at 21:57
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    The measure of any open interval is translation invariant, and these intervals determine Lebesgue measure. – zhw. Mar 21 '16 at 21:57
  • What zhw said is the general moral for basic measure theory problems as well. Proving the theorem in full generality is quite difficult, so one usually reduces to a simple case and realizes that nice approximations are available. In this case, we prove it for open intervals and approximate the measure of any Lebesgue measurable set from there. – user217285 Mar 21 '16 at 22:37
  • @Tryss : only of the Haar measure (invariant for the action of a group, here the group of translations), no there are many measures which are not invariant by translation, for example $\mu([a,b]) = \frac{1}{\sqrt{\pi}} \int_a^b e^{-x^2}dx$ (any probability measure is a measure), there is also the Dirac measure $\mu([a,b]) = 1_{ 0 \in [a,b]} $ (such that $d\mu = \delta(x) dx$), etc. – reuns Mar 21 '16 at 23:33
  • @user1952009 : I meant that the OP said "so far all I have gotten is that $\lambda(A) = \lambda(x+A)$ for all Borel sets $A$". That meant that he has gotten, by definition of a translation invariant mesure, that $\lambda$ is a translation invariant mesure. I know that most measure are not translation invariant – Tryss Mar 22 '16 at 00:32

2 Answers2

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Define $f(x+y)=y, \:y\in A$. Clearly $f$ is continuous and so Borel. Thus $f^{-1}(A)$ is Borel set if A is Borel set.

Since $\{x:f(x+y)<c\}=\{x:f(x)<c-y\}$, $f$ is also Lebesgue measurable.

Eugene Zhang
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For translation invariance:

Let $E \subset \Bbb{R}$ be Lebesgue measurable. Then if the measure is infinite the case is obvious, so suppose its finite and

$$\lambda(E):=\inf\{\sum_{n \in \Bbb{N}}l((a_n,b_n)): E \subset \cup_{n \in \Bbb{N}}(a_n,b_n): l((a_n,b_n))=b_n-a_n\},$$

exists. So we have that

$$E \subset \bigcup_{n \in \Bbb{N}}(a_n,b_n).$$

Then for any $x \in \Bbb{R}$, one has (and this is clear to see set theoretically):

$$E+x \subset \bigcup_{n \in \Bbb{N}}(a_n+x,b_n+x).$$

But then,

\begin{align} l((a_n+x,b_n+x))&=b_n+x-(a_n+x)\\ &=b_n+x-a_n-x\\ &=b_n-a_n\\ &=l((a_n,b_n)). \end{align}

Thus the lengths of the intervals covering $E$ and $E+x$ are the same so taking an infimum yield the same result forcing $\lambda(E)=\lambda(E+x)$ as needed.

Also, this follows if you cover $E$ by unions of $(a_n,b_n]$ or $[a_n,b_n]$ or $[a_n,b_n)$ as these all have same Lebesgue measure as singletons have Lebesgue measure zero.

homosapien
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