Is $\lambda_2(B\times\mathbb{R})$ on the sigma algebra $\sigma:=\{B \times \mathbb{R}:B\in\mathcal{B}\}$ sigma-finite?

Question

$\mu(B \times \mathbb{R}):=\lambda_2(B\times\mathbb{R})$ on the sigma algebra $\sigma:=\{B \times \mathbb{R}:B\in\mathcal{B}\}$. $\lambda_2$ denotes the 2-dimensional Lebesgue measure.

According to my script $\mu$ is not sigma finite.

Can anybody tell me why it is not sigma finite?

The problem I have with this task is that there is no $\Omega$ given.

Thanks in advance!

Answer 1

Here $\Omega=\mathbb R\times\mathbb R$.

If $A\times\mathbb R$ is an elements of the $\sigma$-algebra with $\mu(A\times\mathbb R)<\infty$ then $\lambda_1(A)=0$.

Consequently if $(A_n\times\mathbb R)_n$ is a sequence of measurable sets with finite measure then: $$\lambda_1\left(\bigcup_{n=1}^{\infty}A_n\right)\leq\sum_{n=1}^{\infty}\lambda_1(A_n)=0\text{ hence }\bigcup_{n=1}^{\infty}A_n\neq\mathbb R$$

Then: $$\bigcup_{n=1}^{\infty}(A_n\times\mathbb R)=\left(\bigcup_{n=1}^{\infty}A_n\right)\times\mathbb R\neq\mathbb R\times\mathbb R$$

showing that $\mu$ is not $\sigma$-finite.

Answer 2

To cast away your problem, recall that $\Omega$ is just the biggest element of the $\sigma$-algebra, so in this case $\Omega = \Bbb R^2$. By definition of $\sigma$-finite measures, there should exist an increasing sequence of $A_n\subset \Omega$ such that $\lambda_2(A_n) < \infty$ and the union of $A_n$ is $\Omega$. In your case, do you have any set in your $\sigma$-algebra of non-zero finite measure?