Is Lebesgue measure translation invariant?

Question

I am trying to prove that the Lebesgue measure is translation-invariant. Namely, given a set $X\subseteq\mathbb{R}$, I'd like to show $X + y$ is measurable and $\mathit{m}(X + y) = \mathit{m}(X)$. Namely, that the measures -- not the outter measures alone -- agree. I am mostly stuck on demonstrating that the translation $X + y$ is measurable to begin with. Any ideas?

Answer 1

Definitions

The proof follows from the definitions:

1. For $X \subseteq \mathbb{R}$, the outer measure of set $X$ is defined as $$ m^*(X) := \inf \sum_n l(I_n),$$ where $\{I_n\}_{n \in \mathbb{N}}$ is a countable collection of open intervals such that $X \subset \bigcup_n I_n $ and $l(I_n)$ is the length of $I_n$.

2. If $X \subseteq \mathbb{R}$ is measurable, i.e., $$\forall A \subseteq \mathbb{R}, \quad m^*(A) = m^*(A \cap X) + m^*(A\cap X^c),$$ the Lebesgue measure of $X$ is defined as its outer measure, i.e., $$ m(X) := m^*(X). $$ In other words, the Lebesgue measure is the outer measure restricted to the measurable sets.

Strategy

In order to prove that the Lebesgue measure is translation invariant, we need to prove two things:

1. If $X$ is measurable, then for any $y \in \mathbb{R}$, $X + y = \{x + y: x \in X \}$ is measurable.
2. $m^*(X) = m^*(X+y)$.

Proof

We start by proving the point 2.

Let $\{I_n\}_{n \in \mathbb{N}}$ be a countable collection of open intervals such that $X \subset \bigcup_n I_n$ and $I_n + y= \{z + y: z \in I_n \}$. First, notice that $l(I_n) = l(I_n + y)$ and $X + y \subset \bigcup_n (I_n + y)$. On the one hand, by the definition of the outer measure, $$ m^*(X) \leq \sum_n l(I_n) = \sum_n l(I_n + y),$$ but $ m^*(X+y) = \inf \sum_n l(I_n+y)$, we have $$ m^*(X) \leq m^*(X + y).$$ On the other hand, $$ m^*(X+y) \leq \sum_n l(I_n + y) = \sum_n l(I_n),$$ but $m^*(X) = \inf \sum_n l(I_n)$, we have $$ m^*(X + y) \leq m^*(X).$$ Therefore, $$m^*(X) = m^*(X+y).$$

Let us now prove the point 1.

Let $A \subseteq \mathbb{R}$ and suppose $X$ measurable. First, notice that $(X + y)^c = \{x + y: x \notin X\} = X^c + y$ and $(A + y) \cap (X + y) = (A\cap X) + y$. Then, $$ \begin{align} m^*((A + y) \cap (X + y)) + m^*((A + y) \cap (X + y)^c) &= m^*((A\cap X) + y) + m^*((A\cap X^c) + y) \\ &= m^*(A\cap X) + m^*(A\cap X^c) \\ &= m^*(A). \end{align}$$ The second-last line follows from the point 2 we have just proved, while the last line follows from the measurability of $X$. Therefore, we have $$ m^*(A + y) = m^*(A) = m^*((A + y) \cap (X + y)) + m^*((A + y) \cap (X + y)^c),$$ which means $X + y$ is measurable.

This completes the proof.

Answer 2

Theorem: If $X\subset \Bbb{R}$ is Lebesgue measurable, then for every $y\in \Bbb{R}$, $m(X)=m(X+y)$.

Here $m$ denotes Lebesgue measure.

Proof:

Let $X \subset \Bbb{R}$ be a Lebesgue measruable set, then

$$m(X):=\inf\{\sum_n l((a_n,b_n]): X \subset \cup_n (a_n,b_n]\}.$$

Where $l((a_n,b_n))=b_n-a_n,$ denotes length.

I.e.,

$$X \subset \bigcup_n (a_n,b_n].$$

then for any $y \in \Bbb{R}$, one has

$$X+y \subset \bigcup_n (a_n+y,b_n+y].$$

As $m$ is lebesgue measure, we have that

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

Thus their infimums agree and therefore $m(X+y)=m(X)$, as was needed. $\blacksquare$

Also, for the proof of showing $X+y$ is measurable, refer to @swan11’s answer. Furthermore, for any $a\in \Bbb{R}$, one has

$$m(aX)=\vert a \vert m(X).$$

To see this consider cases were $a\in \Bbb{R}^+$ and $a=0$ and $a\in \Bbb{R}^-$.

Added for future viewers: I used the definition of Lebesgue measurable found in Bass' text, but this works if we cover $X$ by any countable union of intervals, be it $(a_n,b_n),[a_n,b_n],[a_n,b_n)$, this is because the lengths of these intervals has the same Lebesgue measure because singletons have zero Lebesgue measure.

Answer 3

The Lebesgue measure is defined in terms of some basic sets, the open intervals, for example. The measure of a translated open set is the same as the measure of the set, so this structural property carries through to the Lebesgue measure.

That is, if $A$ is a set and $U_k$ forms an open cover by intervals, then $U_k+ \{x\}$ forms an open cover of $A+ \{x\}$. Since the length of $U_k$ and the length of $U_k+\{x\}$ is the same, then just applying the definition shows that $mA = m (A + \{x\})$.

The result is true in $\mathbb{R}^n$ of course.

This sort of approach is used a lot in measure theory.

Answer 4

Observe that the open sets on $\mathbb{R}$ are invariant under translations. That is, for any open $S \subseteq \mathbb{R}, S + y$ is open. Thus the Borel sets are invariant under translations, too. This establishes the measurability.

Now define $m^y(S) = m(S + y)$. We wish to show that $m^y = m$. By definition, Lebesgue measure is generated by the premeasure $m_0$ on the algebra of intervals. Then clearly $m_0 = m_0^y$, the translated premeasure. Then, observe that $\mathbb{R}$ is the countable union of sets with finite measures, namely $\mathbb{R} = \bigcup_{j=-\infty}^\infty [j,j+1)$, so $\mathbb{R}$ is $\sigma$-finite.

The extension of premeasure to a measure is unique when the space is $\sigma$-finite, so $m^y = m$. Finally, we need to show that Lebesgue null sets are preserved by translation. From the conclusion above, any Borel set $S$ satisfies $m(S) = m^y(S)$, and this remains true for sets with zero measure. By completeness any null set is measurable, and the proof is complete.