Proving the scaling property of the Lebesgue measure using Dynkin's lemma

Let $\lambda_1$ be the Lebesgue measure on $(\mathbb{R},\mathcal{B})$ where $\mathcal{B}$ is the Borel $\sigma$-algebra on $\mathbb{R}$.

I would like to show that $\lambda_1$ has the scaling property, i.e. $\lambda_1(cB) = |c|\lambda_1(B)$, for any $B \in \mathcal{B}$ and non-zero $c \in \mathbb{R}$.

I know this can be shown using the uniqueness theorems for measures; I am curious if and how it can be done using the following theorem:

Let $X$ be a non empty set. If $\mathcal{D}$ is a $\pi$-system in $X$, $\mathcal{L}$ is a $\lambda$-system in X and $\mathcal{D} \subset \mathcal{L}$, then $\sigma(\mathcal{D}) \subset \mathcal{L}$.

My idea:

Define $\mathcal{L} := \{A \in \mathcal{B} : \lambda_1(cB) = |c|\lambda_1(B)$ for non-zero $c \in \mathbb{R}\}$

$\mathcal{I} := \{(a,b]: -\infty<a<b\leq\infty \}$

I know that $\mathcal{I}$ is a $\pi$-system and that $\mathcal{I} \subset \mathcal{L}$.

So I just need to show that my $\mathcal{L}$ is a $\lambda$-system, whence $\mathcal{B} = \sigma(\mathcal{I}) \subset \mathcal{L}$ and the scaling property holds true for all Borel sets.

$\mathcal{L}$ is a called a $\lambda$-system if:

(i) $\emptyset \in \mathcal{L}$

(ii) $a \in \mathcal{L} \implies A^{c} \in \mathcal{L}$

(iii) If $A_1,A_2,... \in \mathcal{L}$ are pairwise disjoint then $\cup_{n=1}^{\infty}A_n \in \mathcal{L}$

I believe (i) and (iii) are true but I am not sure if (ii) is true.

Question: Is (ii) above true for my $\mathcal{L}$? If so, how would one show that? If not, what is a counter example?

Solution 1:

Yes, (ii) is definitely true, because of the scaling property of Lebesgue measure. However, proving (ii) directly isn't easy because we're dealing with an infinite measure. The typical way of dealing with this is to exploit $\sigma$-finiteness of Lebesgue measure.

What I would suggest is to define for each $k\in\Bbb{Z}$, $E_k:=[k,k+1)$, and let $\mathcal{L}_k$ the set of all Borel measurable $A\subset E_k=[k,k+1)$ such that $\lambda(cA)=|c|\lambda(A)$. It is immediate that the intervals $\mathcal{I}=\{[a,b)\,:\, k\leq a<b<k+1\}$ is a $\pi$-system on $E_k$ and it is contained in $\mathcal{L}_k$. Points (i) and (iii) hold for very trivial reasons. For point (ii), let $A\in\mathcal{L}_k$ be arbitrary. Then, \begin{align} \lambda(c\cdot (E_k\setminus A))&=\lambda(cE_k)-\lambda(cA)\\ &=|c|\lambda(E_k)-|c|\lambda(A)\\ &=|c|\lambda(E_k\setminus A) \end{align} The subtraction is valid due to the sets having finite measure. Hence, $E_k\setminus A\in\mathcal{L}_k$, and thus $\mathcal{L}_k$ is a $\lambda$-system on $E_k$. By Dynkin's theorem, $\mathcal{L}_k$ equals all Borel subsets of $E_k=[k,k+1)$.

Now, we don't have to worry about proving $\mathcal{L}$ is a $\lambda$-system. We can directly show that it equals all Borel sets because for any Borel set $A\subset\Bbb{R}$, we have \begin{align} \lambda(c\cdot A)&=\lambda\left(c\cdot \bigcup_{k\in \Bbb{Z}}E_k\cap A\right)\\ &=\lambda\left(\bigcup_{k\in\Bbb{Z}}c\cdot (E_k\cap A)\right)\\ &=\sum_{k\in\Bbb{Z}}\lambda\left(c\cdot (E_k\cap A)\right)\\ &=\sum_{k\in\Bbb{Z}}|c|\lambda(E_k\cap A)\tag{$*$}\\ &=|c|\lambda(A) \end{align} where we have repeatedly used countable additivity of measures. Also, $(*)$ is because $E_k\cap A$ is a Borel subset of $E_k$, and hence by what we showed above, the formula holds for it. This proves the formula holds for all Borel subsets. (And from here, since the Lebesgue $\sigma$-algebra is the completion, we can deduce the formula holds for all Lebesgue measurable sets as well).

Note by the way that this reduction to the finite measure case is also one of the possible proofs for the "uniqueness" of Lebesgue measure (i.e it is the unique translation-invariant measure defined on Lebesgue $\sigma$-algebra, finite on compact sets, and normalized to 1, i.e measure of $[0,1)$ is $1$).

Also, essentially the proof outline works for the more general situation of proving that in $\Bbb{R}^n$, for any linear $T:\Bbb{R}^n\to\Bbb{R}^n$ and Lebesgue measurable set $A\subset\Bbb{R}^n$, we have $\lambda(T(A))=|\det T|\lambda(A)$.