If one side of $\int f\ d\lambda = \int f\ d\mu - \int f\ d\nu$ exists, does the other one exist as well? Let $\mu$ and $\nu$ be two positive measures, at least one of which is finite, on a measurable space $(X, \mathfrak{A})$. Let $\lambda$ be a signed measure on $(X, \mathfrak{A})$ defined by setting $\lambda = \mu - \nu$. Let $f$ be an extended real-valued $\mathfrak{A}$-measurable function on $X$.
If $f$ is bounded, it can be shown that
$$
\int_X f\ d\lambda = \int_X f\ d\mu - \int_X f\ d\nu \tag{*}
$$
in the sense that both sides of the equation exist in $\overline{\mathbb{R}} = \mathbb{R} \cup \{\pm \infty\}$ and are equal.
Moreover, if $f$ is unbounded and both sides of $(*)$ exist, it can be shown that the equality holds.
Suppose $f$ is unbounded. Is it possible for only one side of $(*)$ to exist, but not the other? It may be assumed that $f$ is non-negative.

The following theorem from Yeh's "Real Analysis", 2nd edition, claims that the existence of one side implies that of the other one. However, in my opinion, the proof does not prove this; it only proves the two cases mentioned above.
 A: Simpler example: $X=[0,1]$, $\mu=\nu=$ Lebesgue measure on $[0,1]$, $f(t)=1/t$. Then $\lambda=0$ so the left side exists, but the right side is $\infty-\infty$.
A: Let us consider the measurable space $\left(\left(0,\infty\right),\mathcal{B}\right)$,
where $\mathcal{B}$ is the Borel $\sigma$-algebra. Let $\mu$ be
the usual Lebesgue measure on this space.
Let $M_{n}:=\left(n,n+2^{-n}\right)$ for $n\in\mathbb{N}$. Note
that the sets $\left(M_{n}\right)_{n}$ are pairwise disjoint with
$\sum_{n}\mu\left(M_{n}\right)<\infty$. Thus, the measure $\nu$
defined by
$$
d\nu:=\sum_{n\in\mathbb{N}}\chi_{M_{n}}\,{\rm d}\mu
$$
is finite ($\nu$ has density $\sum_{n\in\mathbb{N}}\chi_{M_{n}}$
with respect to $\mu$, where $\chi_{M_{n}}$ is the indicator function
of $M_{n}$).
Note that the "difference measure" $\lambda = \mu - \nu$ is given by
$$
d\lambda=\left(1-\sum_{n\in\mathbb{N}}\chi_{M_{n}}\right)\, d\mu.
$$
In particular, any function $h$ supported on the set $\bigcup_{n\in\mathbb{N}}M_{n}$
will have $\int h\, d\lambda=0$ (and the integral converges absolutely).
All that remains is to choose $h$ in such a way that $\int h\,{\rm d}\mu=\infty$
and $\int h\,{\rm d}\nu=\infty$. The second identity is automatic
if $\int h\,{\rm d}\mu=\infty$ (why? use that $h$ is supported in
$\bigcup_{n\in\mathbb{N}}M_{n}$).
Now a valid choice will be
$$
h=\sum_{n\in\mathbb{N}}2^{n}\chi_{M_{n}},
$$
since we have
$$
\int h\,{\rm d}\mu=\sum_{n\in\mathbb{N}}2^{n}\mu\left(M_{n}\right)=\sum_{n\in\mathbb{N}}1=\infty.
$$
BTW: Good catch in noting the incompleteness of the proof!
