If $\mu, \nu$ are measures on $(X,M)$ and $\nu$ is $\sigma$-finite, and $\mu = \nu + \lambda$, then $\lambda$ is unique Let $(X, M)$ be a measurable space. If $\nu$ and $\lambda$ are measures on $(X,M)$, define measure $\mu = \nu + \lambda$ to be $\mu(E) = \nu(E) + \lambda(E)$.
Assume $\mu$ and $\nu$ are measures on $(X,M)$ such that $\mu(A) \geq \nu(A)$ for all $A \in M$.
1) If $\nu$ is $\sigma$-finite, show that if $\lambda$ is a measure on $(X,M)$ such that $\mu = \nu + \lambda$ then $\lambda$ is unique. Also give a counterexample to show that this result is false without $\nu $ being $\sigma$-finite
2) Show that there always exists a $\lambda$ such that $\mu = \nu + \lambda$.
I was thinking that to show $\lambda$ is unique, we need to show $\lambda_1(E) = \lambda_2(E)$ for all $E$ if $\nu(E) + \lambda_1(E) = \nu(E) + \lambda_2(E)$. I thought if the sum is finite, then we are done. If the sum is infinite, we somehow want to show $\nu(E)$ is finite. But then I got stuck, I didn't know how to get this from the $\sigma$-finiteness and $\mu \geq \nu$ conditions. I also had a hard time coming up with counter examples. Can someone provide a solution to this problem? Thanks. 
 A: I think the trick to this question is to realize that
the intuitive answer of defining $\lambda(E)=\mu(E)-\nu(E)$ would
not be well defined as, one might end up with $\infty-\infty$.
Given, $\nu$ is $\sigma-$finite, $\exists F_{i},\ \nu(F_{i})<\infty,\ ,\cup_{i}F_{i}=X$.
Now, we can also formulate a partition $\{E_{i}\}_{i\in\mathbb{N}}$
of $X$ by $E_{1}=\emptyset,\ E_{i}=F_{i}\backslash\cup_{j<i}F_{j}$,
$\mu(E_{i})\leq\mu(F_{i})<\infty$. We will approach the proof by
contradiction. Let there exist measures $\lambda_{1},\lambda_{2}$
which both satisfy $\mu=\nu+\lambda_{i}$, but, they disagree on a
set $E,$ i.e, $\lambda_{1}(E)\ne\lambda_{2}(E)$.
Now, $\lambda_{i}(E)=\lambda_{i}(\cup_{i}(E\cap E_{i}))=\sum_{i}\lambda_{i}(E\cap E_{i})$.
So, the $\lambda_{i}$'s must disagree on atleast one of the $E\cap E_{j}$.
Now, by $\sigma$-finiteness of $\nu$, $\nu(E\cap E_{j})<\infty$.
Hence, $\lambda_{1}(E\cap E_{j})\ne\lambda(E\cap E_{j})$, but, $\nu(E\cap E_{j})+\lambda_{1}(E\cap E_{j})=\nu(E\cap E_{j})+\lambda_{1}(E\cap E_{j})$,
which is a contradiction.
Now, to see why $\sigma-$finiteness of $\nu$ was crucial for our
result, let us define 
\begin{eqnarray*}
\nu(E)=\mu(E) & = & 0\ \ \text{if }E=\emptyset\\
 & = & \infty\ \mathrm{otherwise}.
\end{eqnarray*}
Now, for arbitrary $x\in X$, and $\forall m>0,$ $\mu(E)=\nu(E)+\lambda(E)$
$ $ and
\begin{eqnarray*}
\lambda(E) & = & 0\ \ \text{if }x\notin  E\\
 & = & \infty\ \mathrm{otherwise}.
\end{eqnarray*}
If, X is singleton, then, we would need :
\begin{eqnarray*}
\lambda(E) & = & 0\ \ \text{if }x\notin E\\
 & = & M\ \mathrm{otherwise\  where\ M\in \mathbb{R}}.
\end{eqnarray*}
I am still stuck at part (2).
