Importance of $\sigma$-finiteness for Uniqueness of Measure $dm_f = f dm$

Question

Background

Given a measure $m : X \rightarrow [0,+\infty]$ and a measurable function $f : X \rightarrow [0,+\infty]$, we can define a new measure by integrating $f$ as below:

$$ m_f(E) = \int_E f \, dm $$

I'm not aware of a name for $m_f$, hence the awkward title. My question arises from Exercise #1.2.2 of Tao, "Epsilon of Room, Vol. I", which states the following:

EOR #1.2.2: Let $m$ be $\sigma$-finite. Given two functions $f,g : X \rightarrow [0,+\infty]$, show that $m_f = m_g$ if and only if $f = g$ for $m$-almost every $x$. Give and example to show that this uniqueness statement can fail if $m$ is not $\sigma$-finite.

I am trying to better understand the importance of the $\sigma$-finiteness condition. For reference, here is my proof of the statement.

Proof

The $\impliedby$ direction is easy and true even when $X$ is not $\sigma$-finite. For the $\implies$ direction, we prove the contrapositive.

Let $\mu(X) < +\infty$ and suppose $f$ is not almost-everywhere equal to $g$, that is, $f \neq g$ on a set $E$ of positive but possibly infinite measure.

1. Without loss of generality, we can take $f > g$ on $E$ since the sets $E \cap 1_{f > g}$ and $E \cap 1_{f < g}$ cannot both be null. This condition prevents $g$ from being infinite on $E$.

2. Moreover, we may assume $\mu(E) < +\infty$ is finite because each infinite-measure set in a sigma-finite space necessarily has a subset of positive measure. This ensures $m_g(E) < +\infty$.

3. Then, $m_f(E) = \int_E f \, dm > \int_E g \, dm = m_g(E)$, where it is possible for $m_f(E) = +\infty$ but $m_g(E)$ must be finite.

Counterexample

Let $X = \{ a \}$ be a singleton set and $\mu : \{ \emptyset, X \} \rightarrow [0,+\infty]$ be the measure which assigns $\mu(\emptyset) = 0$ and $\mu(X) = +\infty$. Clearly $X$ is not $\sigma$-finite. Let $f(a) = 1$ and $g(a) = 2$. Then $m_f(X) = m_g(X) = +\infty$, but $f \neq g$.

Questions

1. What is stopping uniqueness from being true when $X$ is not $\sigma$-finite? Like the counterexample suggests, we run into problems when $f$ and $g$ only disagree on sets of infinite measure, because then it is possible that $m_f(E) = m_g(E) = +\infty$. Is this the only problematic case?
2. For which classes of functions is $m_f$ unique even when $X$ is not $\sigma$-finite? What if we restrict $g$ to be integrable? Then for any $E$, we have $m_g(E) < \int_X g dm < +\infty$. This could be too strong an assumption though since integrable functions have $\sigma$-finite support.