Consider a measure space $(X,\mathcal{M},\mu$). Prove that if $\lambda$ is a measure on $\mathcal{M}$ such that $\lambda(B) \geq \mu(B)$, for all $B \in \mathcal{M}$, then there exists a measure $\nu$ on $\mathcal{M}$, such that $\lambda = \mu + \nu$.
Define $\nu$ such that for all $B \in \mathcal{M}$ such that $\lambda(B) = \infty$ we have $\nu(B) = \infty$, and for all $B \in \mathcal{M}$ such that $\lambda(B) < \infty$ we have $\nu(B) = \lambda(B) - \mu(B)$. Clearly for all $B \in \mathcal{M}$ we have $\lambda(B)  = \mu(B) + \nu(B)$. So it suffice to show $\nu$ is a measure on $\mathcal{M}$. Since for all $B \in \mathcal{M}$ we have $\lambda(B) \geq \nu(B)$, it follows that $\nu(B) = \lambda(B) - \mu(B) \geq 0$. Since $\lambda(\emptyset) = 0 < \infty$ it follows that $\nu(\emptyset) = \lambda(\emptyset) - \mu(\emptyset) = \lambda(\emptyset) - 0 = \lambda(\emptyset) = 0$ where the second equality holds from the fact that $\mu$ is a measure and the fourth equality holds from the fact that $\lambda$ is a measure. Now consider an arbitrary countable collection of disjoint set $\{E_j\}_{j = 1}^\infty \in \mathcal{M}$ such that  $\cup_{j = 1}^\infty E_j \in \mathcal{M}$. Suppose $\lambda(\cup_{j = 1}^\infty E_j) < \infty$, it follows that $\nu(\cup_{j = 1}^\infty E_j) = \lambda(\cup_{j = 1}^\infty E_j) - \mu(\cup_{j = 1}^\infty E_j) = \sum_{j = 1}^\infty \lambda(E_j ) - \mu(\cup_{j = 1}^\infty E_j)= \sum_{j = 1}^\infty \lambda(E_j) - \sum_{j = 1}^\infty \mu(E_j) = \sum_{j = 1}^\infty (\lambda(E_j) - \mu(E_j)) = \sum_{j = 1}^\infty \nu(E_j)$ 
were the second equality holds from the fact that $\lambda$ is a measure on $\mathcal{M}$ and the third equality holds from the fact that $\nu$ is a measure on $\mathcal{M}$.
MY PROBLEM: I don't know how to prove if $\lambda(\cup_{j = 1}^\infty E_j) = \infty$ then $\nu(\cup_{j = 1}^\infty E_j) = \sum_{j = 1}^\infty \nu(E_j)$. 
I think in the case where one of the $\lambda(E_i) = \infty$ it suffices to say $\nu(\cup_{j = 1}^\infty E_j) = \lambda(\cup_{j = 1}^\infty E_j) = \sum_{j = 1}^\infty \lambda(E_j) = \infty = \nu(E_i) = \sum_{j = 1}^\infty \nu(E_j)$. But I don't know how to go about proving if $\sum_{j = 1}^\infty \lambda(E_j) = \infty$ that $\sum_{j = 1}^\infty
 \lambda(E_j) - \mu(E_j) = \infty$ as well?
 A: If we let $\nu(B)=\infty$ when $\lambda(B)=\infty$ and $\nu(B)=\lambda(B)-\mu(B)$ if $\lambda(B)<\infty$, then we run into a problem for $\sigma$-additivity, for example if $(B_n)$ is a sequence of disjoint sets such that $\lambda(B_n)=n^{-1}+n^{-2}$ and $\mu(B_n)=n^{-1}$, then $B:=\bigcup_{n\geqslant 1}B_n$ has an infinite $\lambda$-measure hence $\nu(B)$ is infinite but $\sum_{n\geqslant 1}\nu(B_n)$ is finite. Letting  $\nu(B)=0$ when $\lambda(B)=\infty$ does not work for the same reason. 
Instead, it was suggested in the comments to define 
$$
\nu(B):=\sup\left\{\lambda(A)-\mu(A), A\in\mathcal M, A\subseteq B, \mu(A)<+\infty\right\}.
$$
We have to check that $\nu$ does the job. 


*

*$\nu(\emptyset)=0$ since the only subset of $\emptyset$ is $\emptyset$ and 
$\lambda(\emptyset)-\mu(\emptyset)=0$.

*Also, we observe that if $B\subset B'$, then $\nu(B)\leqslant \nu(B')$ because 
every subset of $B$ is also a subset of $B'$.

*We have to check the equality $\lambda(B)=\mu(B)+\nu(B)$. If $\mu(B)$ is infinite, this is trivial since $\lambda(B)$ is also infinite. If $\mu(B)$ is finite, then $\nu(B)=\mu(B)$ hence the equality hold.

*It remains to check the $\sigma$-additivity. Let $\left(B_j\right)_{j\geqslant 1}$ be a sequence of pairwise disjoint elements of $\mathcal M$. First we prove the sub-additivity. Let $B:=\bigcup_{n\geqslant 1}B_n$. Then 
$$
\nu(B)=\sup\left\{\sum_{j\geqslant 1}\lambda(A\cap B_j)-\mu(A\cap B_j), A\in\mathcal M, A\subseteq B, \mu(A)<+\infty\right\}\\\leqslant 
\sum_{j\geqslant 1}\sup\left\{\lambda(A\cap B_j)-\mu(A\cap B_j), A\in\mathcal M, A\subseteq B, \mu(A)<+\infty\right\}
$$ 
and if $A\in\mathcal M$ is such that $A\subseteq B$, then $A\cap B_j\subset B_j$ and has finite $\mu$-measure hence 
$$
\sup\left\{\lambda(A\cap B_j)-\mu(A\cap B_j), A\in\mathcal M, A\subseteq B, \mu(A)<+\infty\right\}\leqslant \nu(B_j),
$$
which proves that $\nu\left(\bigcup_{j\geqslant 1}B_j\right)\leqslant \sum_{j\geqslant 1}\nu(B_j)$. 


For the other direction, in view of the second bullet, we can assume that $\nu(B_j)$ is finite for all $j$. Let $\varepsilon$ be fixed. By definition of $\nu$, we can find $A_j\in\mathcal M$ such that $\mu(A_j)$ is finite and 
$\lambda(A_j)-\mu(A_j)>\nu(B_j)-\varepsilon 2^{-j}$. Let us fix $J\geqslant 1$ (we have to truncate the union because we are not sure that $\bigcup_{j\geqslant 1}\mu(A_j)$ is finite). Using again the second bullet, we derive that 
$$
\nu\left(\bigcup_{j\geqslant 1}B_j\right)\geqslant\nu\left(\bigcup_{j= 1}^JB_j\right).
$$
Now, $\bigcup_{j=1}^J A_j$ belongs to $\mathcal M$, its $\mu$-measure is finite and is contained in  $\bigcup_{j= 1}^JB_j$ hence 
$$
\lambda\left(\bigcup_{j=1}^J A_j\right)-\mu\left(\bigcup_{j=1}^J A_j\right)
\leqslant \nu\left(\bigcup_{j= 1}^JB_j\right).
$$
Since $A_j\subset B_j$ and the $B_j$ are disjoint, we derive that 
$$
\nu\left(\bigcup_{j= 1}^JB_j\right)\geqslant \sum_{j=1}^J\lambda(A_j)-\mu(A_j)
\geqslant\sum_{j=1}^J\nu(B_j)-\varepsilon.
$$
We got in total that for all $J\geqslant 1$ and all $\varepsilon>0$,
$$
\nu\left(\bigcup_{j\geqslant 1}B_j\right)\geqslant\sum_{j=1}^J\nu(B_j)-\varepsilon
$$
hence we can conclude.
