# Is Lebesgue measure translation invariant?

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?

• I am working with $\mathbb{R}$ Commented Feb 14, 2016 at 2:04
• If $X$ is measurable, then since $T_y (x) = x-y$ is a homeomorphism, it follows that $X+\{y\} = T^{-1}(X)$ is measurable. Commented Feb 14, 2016 at 5:48
• I totally buy that. In fact, I wanted to argue that $T_y(X)$ and X are $\mathit{isomorphic}$ to begin with, or that X and it's translation are each isomorphic to the congruent "centered" at the origin, or something along those lines.. However, this is an analysis class, and for justice' sake, the argument ought to be analytic :-) Commented Feb 14, 2016 at 6:00
• Well, you need more than an isomorphism, you need continuity. Commented Feb 14, 2016 at 6:02
• You just need $T_y$ to be measurable, and (assuming we are dealing with Borel sets here) since $T_y$ is continuous it is measurable, Commented Feb 14, 2016 at 6:08

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.

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 $$l((a_n,b_n))=b_n-a_n.$$

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 $$(a_n,b_n),[a_n,b_n],[a_n,b_n)$$.

• So neat. Thanks for this answer. Commented Nov 7, 2023 at 10:18
• @Kadmos np :-) cheers to more maths!! Commented Nov 7, 2023 at 18:01
• I think the same proof works in higher dimensions with the definition of the measure of a cube as the product of the lengths of its sides. Commented Nov 8, 2023 at 0:30
• @AlexOrtiz this is true, its true for $\Bbb{R}^n$ using $m^n$ the $n-$dimensional Lebesgue measure. Commented Nov 8, 2023 at 2:16

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.

• Cool! Thanks. This is more or less what I have. Of course, the intuition is exactly as you put it, but I am uncomfortable with the following -- How does $\mathit{m}(\cup U_k \setminus X) < \epsilon$ imply $\mathit{m}(\cup U_k + y \setminus X + y) < \epsilon$? Commented Feb 14, 2016 at 3:31
• If $I_j$ is a countable collection of open intervals that cover $\cup U_k \setminus X$ then $I_j+ \{y\}$ is a countable open cover of $\cup U_k + \{y\} \setminus X + \{y\}$. It follows that $m(\cup U_k + \{y\} \setminus X + \{y\}) \le m(\cup U_k \setminus X)$. Swapping the sets shows eqality. Commented Feb 14, 2016 at 5:31
• I agree with everything upto (but not including) the conclusion that $m(∪U_k+{y}\setminus X+{y})≤m(∪U_k\setminus X)$. It's not like $∪U_k+{y}\setminus X+{y}\subseteq ∪U_k\setminus X$ ? Commented Feb 14, 2016 at 5:39
• I never claimed your last statement, I'm not sure how you came up with it. The measure is defined as the infimum of the sum of the lengths of a countable collection of open intervals. It is clear that the sum of the lengths of a countable collection of open intervals is equal to sum of the lengths of a countable collection of translates of the open intervals. Furthermore, if you have a cover of $A$, then the translates of the cover (which have the same sum of lengths) is a cover of $A+\{y\}$. The relationship of the measures follows from this. Commented Feb 14, 2016 at 5:46
• Agreed. Thanks. Commented Feb 14, 2016 at 6:01

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.

• You also need to show that translates of a null set are null.
– Ian
Commented Feb 14, 2016 at 2:12
• I am not quite following the argument. Can we take the Borel sets out of it? Commented Feb 14, 2016 at 2:36
• @Joe Shmo Lebesgue measure is the completion of Borel measure. Are you confused by the part which I refer to premeasures? Commented Feb 14, 2016 at 3:00
• Yes, for starters :-) Commented Feb 14, 2016 at 3:32