# Is Lebsegue Measure Translation Invariant?

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## Solution 1

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.

## Solution 2

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.

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### Joe Shmo

Updated on August 19, 2022

• Joe Shmo 3 months

I am trying to prove that the Lebsegue 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?

• Joe Shmo over 6 years
I am working with $\mathbb{R}$
• copper.hat over 6 years
If $X$ is measurable, then since $T_y (x) = x-y$ is a homeomorphism, it follows that $X+\{y\} = T^{-1}(X)$ is measurable.
• Joe Shmo over 6 years
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 :-)
• copper.hat over 6 years
Well, you need more than an isomorphism, you need continuity.
• Joe Shmo over 6 years
I see. So continuity is preserved under translation, why is continuity important here?
• copper.hat over 6 years
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,
• Ian over 6 years
You also need to show that translates of a null set are null.
• Joe Shmo over 6 years
I am not quite following the argument. Can we take the Borel sets out of it?
• Henricus V. over 6 years
@Joe Shmo Lebesgue measure is the completion of Borel measure. Are you confused by the part which I refer to premeasures?
• Joe Shmo over 6 years
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 $\mathit{m}(\cup U_k + y \setminus X + y) < \epsilon$?
• Joe Shmo over 6 years
Yes, for starters :-)
• copper.hat over 6 years
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.
• Joe Shmo over 6 years
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 $∪U_k+{y}\setminus X+{y}\subseteq ∪U_k\setminus X$ ?
• copper.hat over 6 years
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.
• Joe Shmo over 6 years
Agreed. Thanks.