# I Lebesgue measure under orthogonal transofrmation

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1. Jun 8, 2016

### DavideGenoa

Hello, friends! Let us define the external measure of the set $A\subset \mathbb{R}^n$ as $$\mu^{\ast}(A):=\inf_{A\subset \bigcup_k P_k}\sum_k m(P_k)$$where the infimum is extended to all the possible covers of $A$ by finite or countable families of $n$-paralleliped $P_k=\prod_{i=1}^n I_i$\$, where $I_i$ is of the form $[a_{i,k},b_{i,k}]$ or $(a_{i,k},b_{i,k})$ or $(a_{i,k},b_{i,k}]$ or $[a_{i,k},b_{i,k})$, with $a_{i,k}\le b_{i,k}$, whose measure is defined as $$m(P_k):=\prod_{i=1}^n(b_{i,k}-a_{i,k}).$$

I am intuitively inclined to believe that, if $T\in\text{End}(\mathbb{R}^3)$ is a unitary transformation, then $$\mu^{\ast}(A)=\mu^{\ast}(T(A))$$but I have no idea how we can prove it.
Is that so and, if it is, how can it be proved?
I $\infty$-ly thank any answerer!

2. Jun 8, 2016

### Staff: Mentor

Let us assume your operator $T$ is orthogonal instead of unitary, because you consider real vector spaces here. (The complex case is analogue.)
This means its determinant is ± 1. Thus it also doesn't change the volume of parallelepipeds, cubes resp., i.e. $m(P_k) = m(T(P_k)).$ In addition $T$ is an isometric isomorphism. Now therefore $m(P_k) = m(T^{-1}(P_k))$ and
$$\mu^{\ast}(T(A))=\inf_{T(A)\subset \bigcup_k P_k}\sum_k m(P_k) = \inf_{A\subset \bigcup_k T^{-1}(P_k)}\sum_k m(T^{-1}(P_k)) = \inf_{A\subset \bigcup_k Q_k}\sum_k m(Q_k) = \mu^{\ast}(A)$$
with $Q_k = T^{-1}(P_k).$

3. Jun 8, 2016

### micromass

Staff Emeritus
You can say this, but it's not as trivial as it looks.

OP: take a look at Jones "Lebesgue integration on Euclidean space" it proves exactly what you want.

4. Jun 9, 2016

### martinbn

Why not? If the transformation preserves lengths and angles.

5. Jun 9, 2016

### micromass

Staff Emeritus
Because there it is not clear how to prove a relation between lengths, angles and areas if your definitions are not right.
If you want to say in $\mathbb{R}^2$ that the measure of a rectangle is length times width then this is true only for sides parallel to the axes. That is the only thing that follows right away from the definition. Rectangles whose sides are not parallel to the axes have an area whose definition is a lot more complicated. Proving that it actually is length times width is not obvious.

6. Jun 9, 2016

### martinbn

I am not sure I understand. It seems you are thinking of $\mathbb{R}^2$ with a given measure. But that is not the case in the original post, at least not the way I understand it. You start by defining the area of a rectanlge to be length times height for all rectangles, so there is nothing to prove here, it is the definition. Then you define the outer measure of a set by the inf in the first post when it exists.

7. Jun 9, 2016

### micromass

Staff Emeritus
OK, but then you need to prove your area is 1) consistent 2) satisfies the properties of areas you want such as sigma-additivity. Then that is the part that is not clear.

8. Jun 9, 2016

### DavideGenoa

Mmh... I have got a problem understanding this: in the definition of $\mu^{\ast}(A):= \inf_{A\subset \bigcup_k Q_k}\sum_k m(Q_k)$ the $n$-parallelepipeds are of the form $Q_k=\prod_{i=1}^n I_i$ where $I_i$ is a finite interval, i.e. each of their sides is parallel to one of the axis, while $T^{-1}(P_k)$ is not of such a form, in general... What am I missing? I heartily thank you again!

9. Jun 9, 2016

### Staff: Mentor

I just thought $T$ is an isometry preserving angles and lengths and $d(Ta,Tb) = || T(a-b) || = || a-b || = d(a,b)$ plus I looked up the volume of a parallelepiped which is defined by a determinant. Since $T$ can only change the sign, the volume stays the same.
But as @micromass pointed out in post #3 ff there might be a little work to do in order to prove that the volume does not change, that the volume can be computed in the described way, resp.

10. Jun 10, 2016

### DavideGenoa

My little knowledge of linear algebra and the geometry of Euclidean spaces is enough to understand this, but the problem is, as you note, that we should be able to prove that the $n$-parallelepipeds in the definition of $\mu^{\ast}$ can be chosed with their sides not all parallel to the Cartesian axis.

11. Jun 10, 2016

### Staff: Mentor

12. Jun 10, 2016

### Hawkeye18

One of standard ways to prove that Lebesgue measure is invariant under orthogonal transformations, is to use the theorem about uniqueness of translation invariant measure. It states that in $m$ is the standard Lebesgue--Borel measure (i.e.~restriction of what OP calls $\mu^*$ to Borel sets) in $\mathbb R^n$, and $\sigma$ is a translation invariant Borel measure in $\mathbb R^n$ satisfying $0<\sigma(Q)<\infty$ for the unit cube $Q$, then there exists a constant $c$, $0< c < \infty$ such that $\sigma=c m$.

If $T$ is an orthogonal matrix, define the measure $\sigma$ by $$\sigma(A) = m(TA)$$ (on Borel sets $A$). The measure $\sigma$ is clearly translation invariant, so there exists $c$, $0< c < \infty$ such that $$m(TA)=cm(A)$$ for all Borel sets $A$. Since for the unit ball $B$ we have $TB=B$, the constant $c$ must be $1$.

Another standard way of proving the same fact is to prove that for a linear transformation $T$ and Borel set $A$ one has $$m(TA) = | \operatorname{det} T| m(A)$$ by decomposing $T$ into a product of elementary matrices, and analyzing how each elementary matrix changes the volume. This proof is presented, for example in Folland's book.

Both the above proofs give what OP needed for Borel sets $A$. To get it for arbitrary sets $A$, one needs just a small extra step, let me present it here.

By the definition of $\mu^*$, for any $\varepsilon>0$ we can cover $A$ by countably many parallelipeds $P_k$, $k\ge 1$, such that $$\sum_k m(P_k) < \mu^*(A)+\varepsilon.$$ Then $TA$ is covered by rotated parallelipeds $TP_k$, and $m(TP_k)= m(P_k)$ ($P_k$ are Borel sets).

Since $m$ is restriction of $\mu^*$ onto Borel sigma-algebra, then $m(TP_k)=\mu^*(TP_k$), and we get from the definition of $\mu^*$ that each $TP_k$ can be covered by parallelipeds $R_{k,j}$, $j\ge 1$, such that $$\sum_{j} m(R_{k,j}) < m(TP_k) + \varepsilon/2^k = m(P_k) + \varepsilon/2^k .$$ The set $TA$ is then covered by parallelipeds $R_{k,j}$, $j,k\ge 1$, so $$\mu^*(TA) \le \sum_{k,j} m(T_{k,j}) < \sum_k m(P_k) + \sum_{k\ge 1} 2^{-k} \varepsilon \le \mu^*(A) + 2\varepsilon.$$ So we got that for any $\varepsilon >0$ $$\mu^*(TA) <\mu^*(A)+2 \varepsilon .$$ Since this inequality holds for all $\varepsilon >0$ we conclude that $$\mu^*(TA) \le \mu^*(A) .$$ The transpose $T^T$ of $T$ is also an orthogonal matrix, $T^T(TA) =A$, so we get the opposite inequality for free.

13. Jun 14, 2016

### DavideGenoa

In A Guide to Advanced Real Analysis? (I have found a proof of $\mu(TG)=|\det (A)|\mu(G)$ in F. Jones Lebesgue Integration on Euclidean Space for open sets $G$, but I can generalise it to closed sets only, not to Borel sets).

Since I read that $\mu^{\ast}(TA)=|\det (A)|(A)$ for any measurable $A\subset\mathbb{R}^n$, I think it would be interesting to see whether $TA$ is measurable when $A$ is. The definition I know of (Lebesgue) measurable is: if for any $\varepsilon>0$ there is an elementary set $B$, i.e. a union of rectangles, such that $$\mu^{\ast}(A\triangle B)<\varepsilon$$then $A$ is measurable and its Lebesgue measure is $\mu(A)=\mu^{\ast}(A)$. Is $TA$ measurable when $A$ is? If $\det(T)=0$ I think it is because $TA\subset T\mathbb{R}^n$ and $\mu(T\mathbb{R}^n)=0$ (Jones gives a sketch of proof), but if $\det(T)\ne0$ how can we prove it?
I heartily thank you!!!

14. Jun 16, 2016

### Hawkeye18

I meant "Real Analysis: Modern Techniques and Their Applications" by Follnad, it is done there for Borel sets. But it is a pretty standard proof, it is definitely presented in many texts.
The same proof works for Lebesgue measurable sets, if we show that for any Lebesgue measurable $A$ and invertible linear $T$ the set $TA$ is also Lebesgue measurable (exactly your question).

But this is also pretty simple. First, it is easy to show than in the definition of outer measure we can replace covering by parallelepipeds by covering by Borel sets.
It also can be shown from your definition of Lebesgue measurable sets that a set $A$ is Lebesgue measurable if and only if for any $\varepsilon>0$ there exists a Borel set $B$ such that $$\mu^{\ast}(A\triangle B)<\varepsilon.$$ That immediately implies that the Lebesgue measurability is invariant under linear transformations.

Note that in the above reasoning I used the fact that a linear transformation maps the sets of small measure to sets of small measure, which follow from the fact that $m(TB) = |\operatorname{det} B| m(B)$ for Borel sets $B$.

15. Jun 17, 2016

### DavideGenoa

Where in the book? I cannot find it in chapter 1 on measure...
I will try to prove that we can equivalently take Borel sets in the definitions of external measure and measurability.
Thank you a lot!!!

16. Jun 17, 2016

### Hawkeye18

It is in Ch. 2, s. 2.6, Theorem 2.44. It uses integration theory, in particular the Fubini theorem.

Note, that the proof of the invariance of the Lebesgue---Borel measure under orthogonal transformation that I outlined before in post #12, does not need integration theory. The uniqueness (up to a multiplicative factor) of a translation invariant measure can be easily obtained by an approximation reasoning. And to get from the invariance under orthogonal transformations the formula $m(TA) =|\operatorname{det} T|m(A)$ one just need to notice that this formula is trivial for a diagonal $T$, and then apply the singular value decomposition.

17. Jun 19, 2016

### DavideGenoa

Thank you so much!!!
I am having a hard time proving that we can use Borel sets in the definitions of measurability and outer measure... I have posted a separate question here...