Statement. Suppose $n$ and $m$ are measures, with the latter being $\sigma$-finite. The following are equivalent:
- $n \ll m$
- $\forall \epsilon > 0, n(A) < \epsilon$ whenever $m(A)$ is sufficiently small.
Proof. 2. implies 1.: Suppose $m(A) = 0$. Then $m(A)$ is automatically less than the $\delta$ necessary to make $n(A) < \epsilon$, for any $\epsilon$. So $n(A) = 0$.
1. implies 2.: Suppose otherwise. We can choose a sequence of sets $A_k$ such that $m(A_k) \to 0$ but $n(A_k) > \epsilon$ for all $k$. We may assume WLOG that $m(A_k) < 2^{-k}$. Now, $m(\limsup A_k) \leq m(\cup_{j=k} A_j) \leq \sum_{j=k}^\infty 2^{-j}$, which can be made arbitrarily small as you increase k. Hence $m(\limsup A_k) = 0$, so by assumption $n(\limsup A_k) = 0$. But since $n$ is finite, this means $n(\limsup A_k) = \lim n(\cup_{j=k}A_j) \geq \epsilon$, which is a contradiction. $\blacksquare$
Theorem. Suppose $n \ll m$ are measures on $X$ and $m$ is finite. Then $n(A) = \int_A g \text{d} m$ for some measurable $g$.
Proof. Let F be the set of all measurable functions with $f \in F$ implies $\int_A f \text{d} m \leq n(A)$ for all measurable $A$. This is clearly non-empty, as the zero function qualifies. Therefore $c = \sup_{f\in F} \int_X f \text{d}m$ exists, so we can approximate it via some sequence $f_n$; we can assume the sequence is increasing*, and thus that $g = \lim f_n$ is well-defined. By Monotone Convergence, $c = \lim \int_X f_n \text{d}m = \int_X g \text{d}m$. So $\int_X g \text{d}m \leq n(X)$.
Next, note that $n_0(A) = n(A)\, – \int_A g\, \text{d}m$ is an (unsigned) measure. If it’s identically zero, then we’re done. Otherwise we can find some set $A_0$ such that $n(A_0) > \int_{A_0} g\, \text{d}m$. Since $n \ll m$, $m(A_0)$ is positive or else both sides of that inequality are zero. Furthermore, it also follows that $n(X) > \int_{X} g\,\text{d}m$ by additivity and the definition of $g$.
We can now choose some $\epsilon > 0$ sufficiently small that $n_0(A_0) > \epsilon m(A_0)$. This implies $n(X) > \int_X g + \epsilon 1_{A_0}\, \text{d}m >\int_X g \,\text{d}m$. This is impossible, since $g$ was supposed to be the supremum of functions with this property.