[proofplan]
The argument is the finite-measure estimate that the $L^1$ error is bounded by the measure of the whole space times the uniform error. For each index $n$, the definition of the [uniform norm](/page/Uniform%20Norm) gives the pointwise bound $|f_n(x)-f(x)|\leq \|f_n-f\|_\infty$ on $E$. Integrating this bound and evaluating the integral of the constant function gives
\begin{align*}
\int_E |f_n-f|\,d\mu(x)\leq \mu(E)\|f_n-f\|_\infty.
\end{align*}
The right-hand side tends to $0$ because $\mu(E)<\infty$ and $\|f_n-f\|_\infty\to 0$.
[/proofplan]
[step:Bound the pointwise error by the uniform norm]
For each $n\in\mathbb{N}$, define the measurable function $h_n:E\to[0,\infty)$ by
\begin{align*}
h_n(x)=|f_n(x)-f(x)|
\end{align*}
for every $x\in E$, and define the nonnegative real number
\begin{align*}
c_n:=\|f_n-f\|_\infty.
\end{align*}
Since $f_n-f:E\to\mathbb{R}$ is bounded, $c_n<\infty$. By the definition of the supremum norm, for every $x\in E$,
\begin{align*}
h_n(x)\leq c_n.
\end{align*}
[guided]
Fix an index $n\in\mathbb{N}$. We isolate the quantity whose integral appears in the conclusion by defining $h_n:E\to[0,\infty)$ by
\begin{align*}
h_n(x)=|f_n(x)-f(x)|
\end{align*}
for every $x\in E$. The map $h_n$ is $\mathcal{E}$-measurable because $f_n$ and $f$ are $\mathcal{E}$-measurable real-valued functions, subtraction preserves measurability, and the absolute value map $\mathbb{R}\to[0,\infty)$ is continuous.
Next define
\begin{align*}
c_n:=\|f_n-f\|_\infty.
\end{align*}
Since both $f_n$ and $f$ are bounded functions from $E$ to $\mathbb{R}$, the difference $f_n-f:E\to\mathbb{R}$ is bounded, so $c_n$ is a finite nonnegative real number. The purpose of $c_n$ is to replace the variable function $h_n$ by a constant upper bound. By the definition of the uniform norm as a supremum over $E$, every $x\in E$ satisfies
\begin{align*}
|f_n(x)-f(x)|\leq \sup_{y\in E}|f_n(y)-f(y)|.
\end{align*}
Therefore, for every $x\in E$,
\begin{align*}
h_n(x)\leq c_n.
\end{align*}
[/guided]
[/step]
[step:Integrate the constant upper bound]
For each $n\in\mathbb{N}$, define the constant measurable function $k_n:E\to[0,\infty)$ by
\begin{align*}
k_n(x)=c_n
\end{align*}
for every $x\in E$. The previous step gives $h_n\leq k_n$ pointwise on $E$. Both $h_n$ and $k_n$ are nonnegative $\mathcal{E}$-[measurable functions](/page/Measurable%20Functions), so by monotonicity of the integral for nonnegative measurable functions,
\begin{align*}
\int_E h_n\,d\mu(x)\leq \int_E k_n\,d\mu(x).
\end{align*}
Since $k_n$ is the constant function with value $c_n$,
\begin{align*}
\int_E k_n\,d\mu(x)=c_n\mu(E).
\end{align*}
Thus, for every $n\in\mathbb{N}$,
\begin{align*}
0\leq \int_E |f_n-f|\,d\mu(x)\leq \mu(E)\|f_n-f\|_\infty.
\end{align*}
[guided]
Fix $n\in\mathbb{N}$. The preceding step produced a finite constant $c_n=\|f_n-f\|_\infty$ satisfying $h_n(x)\leq c_n$ for every $x\in E$. To integrate this pointwise bound, define the constant measurable function $k_n:E\to[0,\infty)$ by
\begin{align*}
k_n(x)=c_n
\end{align*}
for every $x\in E$. Since $c_n\geq0$, the function $k_n$ is nonnegative, and it is $\mathcal{E}$-measurable because every constant real-valued function on a measurable space is measurable.
The pointwise inequality from the first step says $h_n\leq k_n$ on $E$. The monotonicity property of the integral for nonnegative measurable functions applies because both functions are nonnegative and $\mathcal{E}$-measurable. Therefore
\begin{align*}
\int_E h_n\,d\mu(x)\leq \int_E k_n\,d\mu(x).
\end{align*}
The right-hand side is the integral of a constant over the whole space. Since $\mu(E)<\infty$ and $k_n$ has value $c_n$, we have
\begin{align*}
\int_E k_n\,d\mu(x)=c_n\mu(E).
\end{align*}
Substituting the definitions of $h_n$ and $c_n$ gives, for every $n\in\mathbb{N}$,
\begin{align*}
0\leq \int_E |f_n-f|\,d\mu(x)\leq \mu(E)\|f_n-f\|_\infty.
\end{align*}
[/guided]
[/step]
[step:Pass to the limit by squeezing]
By hypothesis,
\begin{align*}
\|f_n-f\|_\infty\to 0.
\end{align*}
Since $\mu(E)<\infty$, multiplication by the fixed finite constant $\mu(E)$ gives
\begin{align*}
\mu(E)\|f_n-f\|_\infty\to 0.
\end{align*}
The estimate from the previous step gives, for every $n\in\mathbb{N}$,
\begin{align*}
0\leq \int_E |f_n-f|\,d\mu(x)\leq \mu(E)\|f_n-f\|_\infty.
\end{align*}
Therefore, by the squeeze principle for real sequences,
\begin{align*}
\int_E |f_n-f|\,d\mu(x)\to 0.
\end{align*}
This is the desired conclusion.
[guided]
The previous step reduced the theorem to a statement about real sequences. Define $a_n:=\int_E |f_n-f|\,d\mu(x)$ and $b_n:=\mu(E)\|f_n-f\|_\infty$ for each $n\in\mathbb{N}$. The estimate already proved says
\begin{align*}
0\leq a_n\leq b_n
\end{align*}
for every $n\in\mathbb{N}$.
By hypothesis, $\|f_n-f\|_\infty\to0$. Since $\mu(E)<\infty$, the number $\mu(E)$ is a fixed finite constant, so multiplication by $\mu(E)$ preserves convergence to $0$:
\begin{align*}
b_n=\mu(E)\|f_n-f\|_\infty\to0.
\end{align*}
The squeeze principle for real sequences now applies to $0\leq a_n\leq b_n$ and yields
\begin{align*}
a_n\to0.
\end{align*}
Unpacking the definition of $a_n$, this is exactly
\begin{align*}
\int_E |f_n-f|\,d\mu(x)\to 0.
\end{align*}
Thus [uniform convergence](/page/Uniform%20Convergence) in the stated uniform norm controls the integrals on the finite-[measure space](/page/Measure%20Space) $(E,\mathcal{E},\mu)$.
[/guided]
[/step]
