# Thread: Questions about contents on an algebra of set X

1. ## Questions about contents on an algebra of set X

Hi!

I am totally new to measure theory, so bear in mind that all concepts like $\displaystyle \sigma-algebra$ and $\displaystyle contents$ are new for me.
Problem:
Suppose $\displaystyle A$ is an algebra of subsets on $\displaystyle X$ and $\displaystyle \mu$ and $\displaystyle \nu$ are two contents on $\displaystyle A$ such that $\displaystyle \mu \leq \nu$ and $\displaystyle \mu(X) = \nu(X) < \infty$. Prove that $\displaystyle \mu = \nu$.

I´m not sure how to start.

2. Well if you mean by contents concentration or measure, then I think it's easy.
Assume $\displaystyle \nu(B)> \mu(B)$, this is for every B in the algebra A of X, A must include X as a subset if it's a sigma algebra (because this is an algebra which is closed to finite intersection of subsets of X, and also countable union if I remember correcly), which is a contradiction to the fact that $\displaystyle \nu(X)=\mu(X)$.

3. Okay, so I choose $\displaystyle B = X$, but we have that $\displaystyle \mu(X) = \nu(X)$, so we have a contradiction?

4. Yes.