Define

\[ \mathcal{F} = \Set*{ f: X \to [0, \infty) }{ f \text{ measurable and } \int _A f d\mu \leq \nu (A) }. \]

The function $f \equiv 0$ is in $\mathcal{F} $, since it is a measurable simple function whose integral over every measureable set is zero.

If $f_1$ and $f_2$ are in $\mathcal{F} $, then $f_1 \vee f_2$ is in $\mathcal{F} $. To check, let $A \in \mathcal{A} $, and define the sets $A_1 = \Set{x \in A}{f_1(x) > f_2(x)}$ and $A_2 = \Set{x \in A}{f_1(x) \leq f_2(x)}$. $A_1$ and $A_2$ partition $A$, so

\[ \begin{aligned} \int _A f_1 \vee f_2 &= \int _{A_1} f_1 \vee f_2 + \int _{A_2} f_1 \vee f_2 \\ &= \int _{A_1} f_1 + \int _{A_2} f_2 \\ &\leq \nu (A_1) + \nu (A_2) \end{aligned} \]

Since $A_1$ and $A_2$ partition $A$,

\[ \nu (A_1) + \nu (A_2) = \nu (A_1 \cup A_2) = \nu (A). \]

Select a sequence of functions $\seq{f}$ in $\mathcal{F} $ so that

\[ \lim_{n} \int \seqt{f} = \sup\Set*{\int f}{f \in \mathcal{F} }. \]

Toward ensuring the sequence is increasing, define $g_1 = f_1$, $g_2 = g_1 \vee f_2$, and $g_n = g_{n-1} \vee f_n$ for $n \geq 3$. Using the observation in the previous paragraph, $g_n \in \mathcal{F} $ for each $n$.

Let $g$ be the pointwise limit of the $\seq{g}$. The monotone convergence of integrals shows

\[ \int _A g = \lim_n \int _A g_n. \]

for each $A \in \mathcal{A} $. Since $\int _A g_n \leq \nu (A)$, so too is the limit and thus so too is $\int _A g$. Thus, $g \in \mathcal{F} $. By construction, for $A = X$, $\int g = \sup\Set*{\int f}{f \in \mathcal{F} }$. We have constructed an element of $\mathcal{F} $ attaining the supremum.

We know that the integral of $g$ on $A$ with respect to $\mu $ is bounded above by $\nu (A)$. We want the gap to be zero. Regardless of the gap, the function $\nu _0: \mathcal{A} \to [0, \infty)$ defined by

\[ \nu _0(A) = \nu (A) - \int (g, A, \mu ), \]

for each $A \in \mathcal{A} $ is a positive measure. If $\nu _0$ is identically zero, then there is no gap.

Suppose there is a gap: then there exists a measurable set with strictly positive measure under $\nu _0$. Since the base set contains this set, and measures are monotone, the base set must have stricty positive measure. Since $\mu $ is finite, there exists a natural number $n$ so that

\[ \nu _0(X) > \frac{1}{n}\mu (X). \]

Define a new measure $\nu _1 = \nu _0 - \frac{1}{n}\mu $. Denote a signed-set decomposition of $\nu _1$ by $(P, N)$. Then $\nu _1(A \cap P) \geq 0$, or equivalently,

\[ \nu _0(A \cap P) - \frac{1}{n}\mu (A \cap P) \geq 0, \]

for all $A$, and so

\[ \begin{aligned} \nu (A) &= \nu _0(A) + \int (g, A, \mu ) \\ &\geq \nu _0(A \cap P) + \int (g, A, \mu ). \end{aligned} \]