Difference between two measures and the existence of a smaller one. Let $(X,M)$ be a measurable space and $\mu, \nu$ two measures on $M$ such that $\mu\geq \nu$,I want to show that exists a measure $\lambda$ such that $\mu=\nu+\lambda$. After that, I want to show that even when $\lambda$ is not unique, there is a smaller one.
Searching in other posts and talking to people, I've come to the possible definition $\lambda(E) = \sup\{\mu(A)-\nu(A)\mid A\in M, A\subset E, \nu(A)<\infty\}$.
By monotonicity of $\mu$ and $\nu$ it's easy to see that this definition match the difference $\mu(E)-\nu(E)$ whenever $\nu(E)<\infty$, so I think that's the right one. So I was trying to prove that $\lambda$ was in fact a measure and I'm having trouble in the $\sigma$-additivity:
Let $\{E_j\}_{j\in\mathbb{N}}\subset M$ be a pairwise disjoint family of measurable sets, so for a given \epsilon>0 and for each $j\in\mathbb{N}$, exists $A_j\subset E_j$ measurable with $\nu(A_j)<\infty$ and $\lambda(E_j)<\mu(A_j)+\nu(A_j)+\frac{\epsilon}{2^j}$, I was thinking to use $\cup_{j\in\mathbb{N}}A_j$ to calculate $\lambda(\cup_{j\in\mathbb{N}}E_j)$ but this union may not have finite $\nu$ measure, so... how can I solve this?
After that, I have to show that may be more than one possible $\lambda$, but always will exist a smaller one, I think that the smaller will be that one defined above, but have no clue of how to prove it.
 A: I have a hypothesis but I am not sure whether it is valid or not. See if it helps.
For a collection of distinct measurable set $\{ E_j \} _{j = 1} ^\infty$ and $E = \bigcup _{ j = 1 } ^\infty E_j$, define the set
$$
  S_E = \{ \mu (W) - \nu (W) : W \in M \, , W \subset E \, , \nu(W) < \infty \} \, .
$$
I assume you can show that $\lambda (E) \leq \sum _{j = 1} ^\infty \lambda (E_j)$. For the other direction, following your argument, fix $\epsilon > 0$. For each $j$, we can find measurable $A_j \subset E_j$, such that
$$
  \nu (A_j) < \infty \quad \text{and} \quad 
  \lambda(E_j) - \frac{\epsilon}{2^j} < \mu (A_j) - \nu (A_j) \, .
$$
Summing over $j$ gives
$$
  \sum _{j = 1} ^\infty \lambda (E_j) - \epsilon
  < \sum _{j = 1} ^\infty [ \mu (A_j) - \nu (A_j) ] \, .
$$
Define, for each $j$,
$$
  B_j = \bigcup _{n = 1} ^j A_n \, ,
$$
so that for every $j$, $B_j \in M$, $B_j \subset E$ and $\nu(B_j) < \infty$. By the countable additivity of $\mu$ and $\nu$, simply taking the limit superior over $j$ yields,
$$
\begin{align*}
  \limsup _{j} \ [\mu (B_j) - \nu (B_j)] 
  &= \limsup _{j} \ \sum _{k = 1} ^j [\mu(A_k) - \nu (A_k)] \\
  &= \sum _{k = 1} ^\infty [\mu(A_k) - \nu (A_k)] \\
  &> \sum _{j = 1} ^\infty \lambda (E_j) - \epsilon \, .
\end{align*}
$$
Note that no continuity of measure is used here, we only take a limit of two sequences that is term-wise equal. Since $\{ \mu (B_j) - \nu (B_j) \} _{j = 1} ^\infty$ is a sequence in the set $S_E$,
$$
  \lambda (E) = \sup S_E \geq \limsup _{j} \ \{ \mu (B_j) - \nu (B_j) \}
  > \sum _{j = 1} ^\infty \lambda (E_j) - \epsilon \, .
$$
The inequality above is independent of $\epsilon$, so we can take $\epsilon \to 0^+$ and see that
$$
  \lambda (E) \geq \sum _{j = 1} ^\infty \lambda (E_j) \, .
$$
For your second question, I think you mean that $\lambda$ is the smallest measure satisfying the equality. Suppose not, then there is another measure $\gamma$ on $M$ and a set $E \in M$ for which $\gamma(E) < \lambda (E)$. With $\lambda$ as a supremum and $\gamma (E) < \lambda (E)$, there is a set $A \in M$ for which $A \subset E$, $\nu (A) < \infty$ and
$$
  \gamma (E) < \mu(A) - \nu (A) \leq \lambda (E).
$$
Having assumed $\nu (A) < \infty$, in order for $\gamma = \mu - \nu$ on $M$, it must be the case that
$$
  \gamma (A) = \mu (A) - \nu (A) \, .
$$
However, we have
$$
  A \subset E \quad \text{and} \quad \gamma (E) < \mu (A) - \nu (A) = \gamma (A) \, .
$$
Therefore $\gamma$ cannot be a measure as monotonicity is violated. This contradiction shows that $\lambda$ is indeed the smallest measure.
Extra info:
In case you want to find a measure greater than $\lambda$, consider the measure that equals $\infty$ when $\nu = \infty$.
