Sigma finite measure i'm doing an exercise and i'm struggling to conclude. It is has followed :
Let $(\mathcal{X},\mathcal{A}, \mu         )$ be a measured space.
We define $\nu$ by:
$ \forall A \in \mathcal{A}, \nu(\mathcal{A}):= $ Sup $ \left\{  \mu(B);B  \in \mathcal{A}, B \subset \mathcal{A} \text{ and } \mu(B)< \infty \right\}$.
I showed that this defines a measure on $(\mathcal{X},\mathcal{A} )$ but i now have to show that if $\mu$ is $\sigma$-finite then $\nu=\mu$.
What i tried is this :
$ \forall A \in \mathcal{A},\exists B_0 \in  \mathcal{A}, \mu(B_0)=  $ Sup $  \left\{  \mu(B);B  \in \mathcal{A}, B \subset \mathcal{A} \text{ and } \mu(B)< \infty \right\}=\nu(A)$
$ B_0 \subset A, \implies \mu(B_0) \leq \mu(A)$.
Then i want to use this
$ \mu(A)= \mu(B_0)+\mu(A \backslash B_0) $ because $\mu(B_0) < \infty$.
There problem i think is that here i want to say that since $\mu(B_0) < \infty$ and $\mu(A) < \infty$ then (i think this is wrong) $\mu(A \backslash B_0)< \infty$.
If this was correct i could say that since $B_0$ realise the supremum of $E= \left\{  \mu(B);B  \in \mathcal{A}, B \subset \mathcal{A} \text{ and } \mu(B)< \infty \right\}$ ,
$\mu(B_0) \leq \mu(A)=\mu(B_0)+\mu(A \backslash B_0) \leq \mu(B_0)+\mu(B_0)=2\mu(B_0) < \infty $.
If this was true we could say that $\mu(A)=\nu(A),  \forall A\in \mathcal{A}$.
I feel like i do not use the fact that $\mu$ is $\sigma$-finite.
The problem is probably that $A \backslash B_0 \notin E$ thus $\mu(A \backslash B_0)$ is not an element we consider here.
Any help would be appreciated !
 A: To show what you want, you can start observing that whenever $\mu(A) < \infty$, the equality holds. Then, you show that it holds for $\sigma$-finite sets, using $\sigma$-additivity. Then it hods for every set if the space is $\sigma$-additive.
I think you should search for an example where the equality fails, so you understand a bit better what is behind the phenomena.
For example, take an uncountable set. For example, $\mathbb{R}$. And take
$$\mu(A) = \begin{cases}0, &\text{ $A$ is countable}\\\infty &\text{ $A$ is uncountable.}\end{cases}$$
Then $\nu = 0$.
A: Suppose $\mu$ is $\sigma$-finite, then we have $X=\bigcup\limits_{n=0}^{\infty} A_n$  with $\forall n \in \mathbb{N}, \mu(A_n) < \infty$ with every $A_n$  two by two disjoint.
$\forall A \in \mathcal{A}, A =\bigcup\limits_{n=0}^{\infty} (A \bigcap  A_n)$,
$\mu(A) =\mu (\bigcup\limits_{n=0}^{\infty} (A \bigcap  A_n)) = \sum\limits_{n=0}^\infty \mu( A \bigcap A_n) \leq \sum\limits_{n=0}^\infty  \mu(A_n)< \infty$ because $\forall n \in \mathbb{N}, \mu(\bigcup\limits_{i=0}^{n} (A \bigcap  A_n))= \sum\limits_{i=0}^{n} \mu(A \bigcap A_n)<\infty$
$   \mu (\bigcup\limits_{n=0}^{\infty} (A \bigcap  A_n)) = \sum\limits_{n=0}^\infty \mu( A \bigcap A_n)       \leq   $ Sup $ \mu (\bigcup\limits_{n=0}^{\infty} (A \bigcap  A_n))=$ Sup $\sum\limits_{n=0}^\infty \mu( A \bigcap A_n) < \infty $. Thus $\mu(A)\leq \nu(A)$.
The other inequality is obvious, since $\nu(A)$ is defined as a supremum.
$\forall A,B \in \mathcal{A}, B \subset A \implies \mu(B) \leq \mu(A) \implies \nu(A) \leq \mu(A)$
Then $\forall A \in \mathcal{A}, \nu(A)=\mu(A) $
Is this correct ?
