[proofplan]
The proof is a direct quantitative version of the usual dominated convergence estimate. For a prescribed error scale $2^{-k}$, we choose the finite-measure set $A_r$ with $r=k+3$, so the tail $E\setminus A_r$ contributes less than one quarter of the allowed error after domination by $2g$. On $A_r$, we split into the exceptional set $B_{n,k,r}$ and its complement: domination controls the exceptional set, while the explicit uniform bound controls the complement using $\mu(A_r)<\infty$. Adding the three estimates gives the computable modulus $M(k)=N(k,k+3)$.
[/proofplan]
[step:Verify that the $L^1$ norms under consideration are finite]
Fix $n\in\mathbb N$. Since $f_n:(E,\mathcal E)\to(\mathbb R,\mathcal B(\mathbb R))$ and $f:(E,\mathcal E)\to(\mathbb R,\mathcal B(\mathbb R))$ are measurable maps, the difference $f_n-f:(E,\mathcal E)\to(\mathbb R,\mathcal B(\mathbb R))$ is measurable. Since $|f_n|\leq g$ and $|f|\leq g$ $\mu$-almost everywhere, there is a measurable null set $Z_n\in\mathcal E$ such that for every $x\in E\setminus Z_n$, the triangle inequality in $\mathbb R$ gives
\begin{align*}
|f_n(x)-f(x)|\leq |f_n(x)|+|f(x)|\leq 2g(x).
\end{align*}
Thus $|f_n-f|\leq 2g$ $\mu$-almost everywhere on every measurable subset of $E$.
Because $g\in L^1(E,\mathcal E,\mu)$, monotonicity of the non-negative integral gives
\begin{align*}
\int_E |f_n(x)-f(x)|\,d\mu(x)\leq 2\int_E g(x)\,d\mu(x)<\infty.
\end{align*}
Thus $f_n-f\in L^1(E,\mathcal E,\mu)$ and $\|f_n-f\|_{L^1(E)}$ is finite for every $n\in\mathbb N$.
[/step]
[step:Choose the finite-measure approximation set from the desired error scale]
Fix $k\in\mathbb N$. Define
\begin{align*}
r:=k+3
\end{align*}
and define the candidate convergence modulus $M:\mathbb N\to\mathbb N$ by
\begin{align*}
M(k):=N(k,k+3).
\end{align*}
Let $n\geq M(k)$. By the hypothesis with the pair $(k,r)$, there is a measurable set $B_{n,k,r}\subset A_r$ such that
\begin{align*}
\int_{B_{n,k,r}} g(x)\,d\mu(x)<2^{-(k+2)}
\end{align*}
and
\begin{align*}
|f_n(x)-f(x)|<2^{-(k+2)}(1+\mu(A_r))^{-1}
\end{align*}
for every $x\in A_r\setminus B_{n,k,r}$.
[/step]
[step:Decompose the $L^1$ error into tail, exceptional, and uniform parts]
The three sets $E\setminus A_r$, $B_{n,k,r}$, and $A_r\setminus B_{n,k,r}$ are measurable and pairwise disjoint, and their union is $E$. Additivity of the integral over this measurable disjoint decomposition gives
\begin{align*}
\|f_n-f\|_{L^1(E)} = \int_{E\setminus A_r} |f_n(x)-f(x)|\,d\mu(x) + \int_{B_{n,k,r}} |f_n(x)-f(x)|\,d\mu(x) + \int_{A_r\setminus B_{n,k,r}} |f_n(x)-f(x)|\,d\mu(x).
\end{align*}
[guided]
We split the integral according to the three regions where different information is available. The set $E\setminus A_r$ is the tail, and the hypothesis says $g$ has small integral there. The set $B_{n,k,r}$ is the exceptional part inside $A_r$, where we do not have a pointwise small bound for $|f_n-f|$ but we do know that $g$ has small integral. The remaining set $A_r\setminus B_{n,k,r}$ is the good part, where the difference $|f_n-f|$ is uniformly small.
The sets involved are measurable: $A_r\in\mathcal E$ by hypothesis, $B_{n,k,r}\in\mathcal E$ by hypothesis, and therefore $E\setminus A_r$ and $A_r\setminus B_{n,k,r}$ are measurable because $\mathcal E$ is a $\sigma$-algebra. They are pairwise disjoint, and since $B_{n,k,r}\subset A_r$, their union is exactly $E$. Hence the additivity of the non-negative integral over disjoint measurable sets yields
\begin{align*}
\int_E |f_n(x)-f(x)|\,d\mu(x) = \int_{E\setminus A_r} |f_n(x)-f(x)|\,d\mu(x) + \int_{B_{n,k,r}} |f_n(x)-f(x)|\,d\mu(x) + \int_{A_r\setminus B_{n,k,r}} |f_n(x)-f(x)|\,d\mu(x).
\end{align*}
This is the structural point of the proof: each of the three summands will be bounded by a specified fraction of $2^{-k}$.
[/guided]
[/step]
[step:Bound the tail and exceptional contributions by domination]
On $E\setminus A_r$, the domination estimate $|f_n-f|\leq 2g$ holds $\mu$-almost everywhere. Since changing an integrand on a $\mu$-null set does not change its non-negative integral, monotonicity of the non-negative integral gives
\begin{align*}
\int_{E\setminus A_r} |f_n(x)-f(x)|\,d\mu(x)\leq 2\int_{E\setminus A_r} g(x)\,d\mu(x).
\end{align*}
Since $r=k+3$, the defining property of $A_r$ gives
\begin{align*}
2\int_{E\setminus A_r} g(x)\,d\mu(x)<2\cdot 2^{-(k+3)}=2^{-(k+2)}.
\end{align*}
On $B_{n,k,r}$, the same almost-everywhere domination estimate and null-set invariance of the non-negative integral give
\begin{align*}
\int_{B_{n,k,r}} |f_n(x)-f(x)|\,d\mu(x)\leq 2\int_{B_{n,k,r}} g(x)\,d\mu(x).
\end{align*}
By the defining property of $B_{n,k,r}$,
\begin{align*}
2\int_{B_{n,k,r}} g(x)\,d\mu(x)<2\cdot 2^{-(k+2)}=2^{-(k+1)}.
\end{align*}
[/step]
[step:Bound the good contribution using finite measure]
On $A_r\setminus B_{n,k,r}$, the pointwise estimate in the hypothesis gives
\begin{align*}
|f_n(x)-f(x)|<2^{-(k+2)}(1+\mu(A_r))^{-1}.
\end{align*}
By monotonicity of the integral and the inclusion $A_r\setminus B_{n,k,r}\subset A_r$,
\begin{align*}
\int_{A_r\setminus B_{n,k,r}} |f_n(x)-f(x)|\,d\mu(x)\leq 2^{-(k+2)}(1+\mu(A_r))^{-1}\mu(A_r).
\end{align*}
Since $\mu(A_r)<\infty$, we have $\mu(A_r)(1+\mu(A_r))^{-1}<1$, and therefore
\begin{align*}
\int_{A_r\setminus B_{n,k,r}} |f_n(x)-f(x)|\,d\mu(x)<2^{-(k+2)}.
\end{align*}
[guided]
The good set is where the proof uses finite measure. The hypothesis gives a uniform pointwise bound on $A_r\setminus B_{n,k,r}$:
\begin{align*}
|f_n(x)-f(x)|<2^{-(k+2)}(1+\mu(A_r))^{-1}.
\end{align*}
Because this upper bound is constant in $x$, integrating it over $A_r\setminus B_{n,k,r}$ gives
\begin{align*}
\int_{A_r\setminus B_{n,k,r}} |f_n(x)-f(x)|\,d\mu(x)\leq 2^{-(k+2)}(1+\mu(A_r))^{-1}\mu(A_r\setminus B_{n,k,r}).
\end{align*}
The inclusion $A_r\setminus B_{n,k,r}\subset A_r$ implies $\mu(A_r\setminus B_{n,k,r})\leq\mu(A_r)$. Hence
\begin{align*}
\int_{A_r\setminus B_{n,k,r}} |f_n(x)-f(x)|\,d\mu(x) \leq 2^{-(k+2)}\frac{\mu(A_r)}{1+\mu(A_r)}.
\end{align*}
The factor $(1+\mu(A_r))^{-1}$ was chosen precisely to make this estimate independent of the size of $A_r$. Since $\mu(A_r)<\infty$, the ratio $\mu(A_r)/(1+\mu(A_r))$ is strictly less than $1$. Therefore
\begin{align*}
\int_{A_r\setminus B_{n,k,r}} |f_n(x)-f(x)|\,d\mu(x)<2^{-(k+2)}.
\end{align*}
[/guided]
[/step]
[step:Add the three bounds to obtain the explicit convergence modulus]
Combining the decomposition with the three estimates gives
\begin{align*}\|f_n-f\|_{L^1(E)} < 2^{-(k+2)}+2^{-(k+1)}+2^{-(k+2)}.\end{align*}
Since
\begin{align*}
2^{-(k+2)}+2^{-(k+1)}+2^{-(k+2)}
=
2^{-k},
\end{align*}
we obtain
\begin{align*}
\|f_n-f\|_{L^1(E)}<2^{-k}.
\end{align*}
Thus for every $k\in\mathbb N$ and every $n\geq M(k)=N(k,k+3)$, the $L^1$ error is less than $2^{-k}$. This proves $\|f_n-f\|_{L^1(E)}\to 0$, with convergence modulus computable from the given data by $M(k)=N(k,k+3)$.
[/step]
