Let $\lambda$ be the lebesgue measure on $\mathbb{R}$, Let $\mathcal{M}(\mathbb{R}^n)$ be the system of lebesgue measurable sets of $\mathbb{R}^n$ and $t>0$.
Show that $\mu: \mathcal{M}(\mathbb{R}^n)\to [0,\infty], \mu(A):=t^{-n}\lambda(tA)$ defines a measure and that $\lambda(A)=\mu(A)$ for all $A\in \mathcal{M}(\mathbb{R}^n)$
My work:
We have:$$\mu(\emptyset)=t^{-n}\cdot \lambda(t\cdot \emptyset)=t^{-n}\cdot \lambda(\emptyset)=t^{-n}\cdot 0=0 \quad \checkmark$$ and $$\mu\left(\bigcup\limits_{n=1}^{\infty}A_n\right)=t^{-n}\lambda \left(\bigcup\limits_{n=1}^{\infty}tA_n\right)\\ =t^{-n}\sum \limits_{n=1}^{\infty}\lambda(tA_n)=t^{-n}\sum \limits_{n=1}^{\infty}t^n\cdot t^{-n}\lambda(tA_n) =t^{-n}\sum \limits_{n=1}^{\infty}t^n\mu(A_n)$$ Hence:$$\mu\left(\bigcup\limits_{n=1}^{\infty}A_n\right)=\sum \limits_{n=1}^{\infty}\mu(A_n) \quad \checkmark $$
All in all, $\mu$ is a measure. But how can I show the second part, namely that $\lambda(A)=\mu(A)$ for all $A\in \mathcal{M}(\mathbb{R}^n)$?
Best Answer
The Lebesgue measure is defined by the restriction of the lebesgue outer measure to lebesgue measurable sets. The outer measure is defined by
$$ \lambda^*(A) = \inf \{ \sum_{j} \text{Vol} (I_j) \; \vert \; A \subset \bigcup_{j \in \mathbb N} I_j \}$$
where the $I_j$'s are compact rectangles which cover $A$.
Let $A$ be a lebesgue measurable set and $t > 0$. We start by showing that $tA$ is lebesgue measurable. To do this we establish the fact that $$ \lambda^*(tA) = t^n \lambda^*(A) $$
This follows by definition of the outer measure
\begin{align*} \lambda^*(tA) &= \inf \{ \sum_{j} \text{Vol} (I_j) \; \vert \; tA \subset \bigcup_{j \in \mathbb N} I_j \} \\ &= \inf \{ \sum_{j} \text{Vol} (I_j) \; \vert \; A \subset \bigcup_{j \in \mathbb N} \frac{1}{t}I_j \} \\ &= \inf \{ t^n \sum_{j} t^{-n}\text{Vol} (I_j) \; \vert \; A \subset \bigcup_{j \in \mathbb N} \frac{1}{t}I_j \} \\ &= \inf \{ t^n \sum_{j}\text{Vol} (\frac 1 tI_j) \; \vert \; A \subset \bigcup_{j \in \mathbb N} \frac{1}{t}I_j \} \\ &= t^n \inf \{\sum_{j}\text{Vol} (\frac 1 tI_j) \; \vert \; A \subset \bigcup_{j \in \mathbb N} \frac{1}{t}I_j \} \\ &= t^n \inf \{\sum_{j}\text{Vol} (\tilde{I}_j) \; \vert \; A \subset \bigcup_{j \in \mathbb N} \tilde{I}_j\} \\ &= t^n \lambda^*(A). \end{align*}
We now show that $tA$ is lebesgue measurable if $A$ is measurable. By definition we must show that for any $E \subset \mathbb R^n$
$$ \lambda^*(E) = \lambda^*(E \cap tA) + \lambda^*(E \cap (tA)^c)$$
Since $(tA)^c = tA^c$ we have
\begin{align*} \lambda^*(E \cap tA) + \lambda^*(E \cap (tA)^c) &= \lambda^*(E \cap tA) + \lambda^*(E \cap tA^c) \\ &= \lambda^*(t \frac 1 t E \cap tA) + \lambda^*(t \frac 1 t E \cap tA^c) \\ &= \lambda^*\big( t (\frac 1 t E \cap A) \big) + \lambda^*\big( t(\frac 1 t E \cap A^c)\big) \\ &= t^n \lambda^*( \frac 1 t E \cap A) + t^n \lambda^*(\frac 1 t E \cap A^c)\\ &= t^n \big( \lambda^*( \frac 1 t E \cap A) + \lambda^*(\frac 1 t E \cap A^c)\big) \\ &= t^n \lambda^*(\frac 1 t E) = \lambda^*(E). \end{align*}
Therefore $tA$ is Lebesgue measurable. Since $tA$ and $A$ are both Lebesgue measurable we have by definition that
$$ \lambda^*(tA) = \lambda(tA) \quad \text{and} \quad \lambda^*(A) = \lambda(A)$$
Hence $ \lambda(tA) = t^n \lambda(A)$ which is equivalent to $\lambda(A) = \mu(A)$.