[proofplan]
We first choose a measurable representative of $g$ and replace it by the everywhere nonnegative function $h:E\to[0,\infty)$ given by $h(x)=\max\{g(x),0\}$, since $h=g$ $\mu$-a.e. and null modifications do not change integrals. This reduces all positivity questions to the case of an everywhere nonnegative integrable function. Countable additivity follows by applying the [Monotone Convergence Theorem](/theorems/509) to the increasing partial sums of the functions $\mathbb 1_{A_n}h$ for a pairwise disjoint family $(A_n)_{n\in\mathbb N}$. Finally, if $\mu(A)=0$, the integral of $h$ over $A$ vanishes, so $\nu_g(A)=0$.
[/proofplan]
[step:Replace the density by an everywhere nonnegative representative]
Choose a measurable representative
\begin{align*}
g:E&\to\mathbb R
\end{align*}
of the given $L^1(E,\mathcal E,\mu)$ class. Define
\begin{align*}
h:E&\to[0,\infty)
\end{align*}
\begin{align*}
x&\mapsto \max\{g(x),0\}.
\end{align*}
Since $g\ge 0$ $\mu$-a.e., we have $h=g$ $\mu$-a.e. Therefore, for every $A\in\mathcal E$,
\begin{align*}
\int_A g\,d\mu=\int_A h\,d\mu.
\end{align*}
Thus $\nu_g(A)=\int_A h\,d\mu$ for all $A\in\mathcal E$. Also $h\in L^1(E,\mathcal E,\mu)$ because $|h|\le |g|$ $\mu$-a.e.
[guided]
The only subtle point is that $g\in L^1(E,\mathcal E,\mu)$ may be regarded as an equivalence class of functions, while the formula defining $\nu_g$ uses a representative. We handle this by choosing one measurable representative
\begin{align*}
g:E&\to\mathbb R.
\end{align*}
Because the hypothesis says $g\ge 0$ $\mu$-a.e., the negative values of this representative can occur only on a $\mu$-null set. Define the nonnegative [measurable function](/page/Measurable%20Function)
\begin{align*}
h:E&\to[0,\infty)
\end{align*}
\begin{align*}
x&\mapsto \max\{g(x),0\}.
\end{align*}
Then $h=g$ outside a $\mu$-null set, hence $h=g$ $\mu$-a.e. The [Lebesgue integral](/page/Lebesgue%20Integral) is unchanged under [modification on a null set](/theorems/4915), so for every measurable set $A\in\mathcal E$,
\begin{align*}
\int_A g\,d\mu=\int_A h\,d\mu.
\end{align*}
This shows that the set function $\nu_g$ can be studied using the everywhere nonnegative density $h$. Since $|h|\le |g|$ $\mu$-a.e. and $g\in L^1(E,\mathcal E,\mu)$, we also have $h\in L^1(E,\mathcal E,\mu)$.
[/guided]
[/step]
[step:Verify positivity and finiteness]
For each $A\in\mathcal E$, the function $\mathbb 1_A h:E\to[0,\infty)$ is measurable and nonnegative, so
\begin{align*}
\nu_g(A)=\int_A h\,d\mu\ge 0.
\end{align*}
Moreover, since $0\le \mathbb 1_A h\le h$ and $h\in L^1(E,\mathcal E,\mu)$,
\begin{align*}
\nu_g(A)=\int_E \mathbb 1_A h\,d\mu\le \int_E h\,d\mu<\infty.
\end{align*}
In particular,
\begin{align*}
\nu_g(\varnothing)=\int_E \mathbb 1_{\varnothing}h\,d\mu=0.
\end{align*}
[/step]
[step:Prove countable additivity on pairwise disjoint measurable sets]
Let $(A_n)_{n\in\mathbb N}$ be a pairwise disjoint sequence in $\mathcal E$, and define
\begin{align*}
A=\bigcup_{n=1}^{\infty}A_n.
\end{align*}
For each $m\in\mathbb N$, define
\begin{align*}
s_m:E&\to[0,\infty)
\end{align*}
\begin{align*}
x&\mapsto \sum_{n=1}^{m}\mathbb 1_{A_n}(x)h(x).
\end{align*}
Each $s_m$ is measurable and nonnegative because it is a finite sum of measurable nonnegative functions.
Because the sets $A_n$ are pairwise disjoint,
\begin{align*}
s_m=\mathbb 1_{\bigcup_{n=1}^{m}A_n}h.
\end{align*}
The sequence $(s_m)_{m\in\mathbb N}$ is increasing pointwise and converges pointwise to $\mathbb 1_Ah$. By the Monotone Convergence Theorem,
\begin{align*}
\int_E \mathbb 1_Ah\,d\mu=\lim_{m\to\infty}\int_E s_m\,d\mu.
\end{align*}
For each $m\in\mathbb N$, finite additivity of the integral gives
\begin{align*}
\int_E s_m\,d\mu=\sum_{n=1}^{m}\int_E\mathbb 1_{A_n}h\,d\mu.
\end{align*}
Therefore
\begin{align*}
\nu_g(A)=\sum_{n=1}^{\infty}\nu_g(A_n).
\end{align*}
Thus $\nu_g$ is countably additive on $\mathcal E$.
[/step]
[step:Show absolute continuity with respect to $\mu$]
Let $A\in\mathcal E$ satisfy $\mu(A)=0$. Since $\mathbb 1_Ah=0$ $\mu$-a.e., the integral of this integrable function is zero:
\begin{align*}
\nu_g(A)=\int_A h\,d\mu=\int_E\mathbb 1_Ah\,d\mu=0.
\end{align*}
Hence $\mu(A)=0$ implies $\nu_g(A)=0$, which is precisely $\nu_g\ll\mu$. Combining positivity, finiteness, $\nu_g(\varnothing)=0$, and countable additivity, $\nu_g$ is a finite positive measure on $(E,\mathcal E)$ absolutely continuous with respect to $\mu$.
[/step]
