I am confused by notes we previously made. Let $\mathcal{B}^{d}$ be the Borel sigma-algebra and $\lambda^{d}$ be the Lebesgue-Borel measure. It was defined as follows:
For all $Q:=\times_{i=1}^{d}[a_{i},b_{i}[$ where $-\infty < a_{i} \leq b_{i} < \infty$ $\forall i = 1,…, d$. It follows that:
$\lambda^{d}(Q)= \prod_{i=1}^{d}(b_{i}-a_{i})$
Let's look at the case $d=1$ and we know $\mathbb R \in \mathcal{B}$ by sheer definition of a $\sigma-$algebra. But surely since
$\lambda(\mathbb R)=\infty$, it cannot be Lebesgue measurable? Or am I confusing definitions here?
Best Answer
A subset $B$ of the set $A$ is measurable or not w.r.t. the sigma algebra $\mathscr A$ on $A$ if and only if $B\in \mathscr A$, we don't need measure.
In case of Lebesgue measure on $\Bbb R^d$ we first define Lebesgue outer measure $\lambda^*:P(\Bbb R^d)\rightarrow [0,\infty]$ as $\lambda^*(A)=$ infimum of of all sums of the form $\sum_{j=1}^{\infty} vol(R_j)$ where $R_j$ is a bounded $d$-dimensional rectangle of the form $\times_{i=1}^{d}[a_{i},b_{i}[$ with $-\infty < a_{i} \leq b_{i} < \infty$ such that $\bigcup_{j=1}^{\infty} R_j\supseteq A$ . Now we define a subset $B$ of $\Bbb R^d$ is Lebesgue measurable if it satisfies the equation $$\lambda^*(A)=\lambda^*(A\cap B)+\lambda^*(A\cap B^c),\forall A\in P(\Bbb R^d).$$ You can show that collection $\mathcal M$ of all Lebesgue measurable subsets of $\Bbb R^d$ is a $\sigma$-algebra and our Lebesgue measure is nothing but $\lambda^*|_{\mathcal M}:\mathcal M\rightarrow [0,\infty]$.