Let $(X,\mathcal{A})$ be a measurable space and $\mu$ and $\nu$ be a signed measures on $(X,\mathcal{A})$. We say that $\lambda =\mu+\nu$ is well-defined as a function $\lambda \: \mathcal{A} \to \overline {\mathbb R}$ given by $\lambda (A) := \mu(A)+\nu(A) \quad \text{for $A\in \mathcal{A}$}$ if the latter sum always makes sense, i.e., iff $\{\mu(A),\nu(A)\}\ne \{-\infty, +\infty\}$ for all $A\in \mathcal{A}$.
Show that $\lambda =\mu+\nu$ is well-defined if and only if both $\mu$ and $\nu$ do not take the value $-\infty$ or both do not take the value $+\infty$. Show that in this case $\lambda: \mathcal{A} \to \overline {\mathbb R}$ is a signed measure.
I have proven the backward implication, and I am having difficulty proving the backward case. Suppose $\mu$ takes $\infty$ and $\nu$ takes $-\infty$, for example. Then $\mu(A) = \infty$ and $\nu(B) = -\infty$ for some $A, B \in \mathcal A$. We want to find some $C \in \mathcal A$ such that $\{ \mu(C), \nu(C) \} = \{ \infty, -\infty \}$. I am not sure how I can find such a $C$.
$\mu(A)=+\infty\implies\mu(X)=+\infty\quad$ and $\quad\nu(B)=-\infty\implies\nu(X)=-\infty$.
So just take $C=X$.
EDIT : for clarity I add some details (essentially what geetha290krm already wrote as a comment below). By definition of a signed measure, if $A$ and $B$ are disjoint measurable sets, then $\mu(A)+\mu(B)$ is well-defined and equal to $\mu(A\cup B)$. Therefore, if $\mu(A)=+\infty$, then $\mu(A)+\mu(A^\complement)$ must be well-defined, which implies that $\mu(A^\complement)\neq-\infty$. Hence $\mu(A)+\mu(A^\complement)=+\infty=\mu(A\cup A^\complement)=\mu(X)$. A similar argument holds to show that $\nu(B)=-\infty\implies \nu(X)=-\infty$.