[proofplan]
We prove the three implications separately from the definitions. Uniform convergence gives a single index after which the error is small at every point, hence in particular at each fixed point. $L^p$ convergence controls the measure of the exceptional set $\{|f_n-f|>\varepsilon\}$ by an integral estimate. Almost everywhere convergence on a finite measure space is converted to convergence in measure by using decreasing tail sets and the continuity from above of finite measures.
[/proofplan]
[step:Specialize the uniform bound to a fixed point]
Assume $f_n\to f$ uniformly on $E$. Let $x\in E$ and let $\varepsilon>0$. By uniform convergence, there exists $N\in\mathbb{N}$ such that for every $n\ge N$ and every $y\in E$,
\begin{align*}
|f_n(y)-f(y)|<\varepsilon.
\end{align*}
Taking $y=x$ gives
\begin{align*}
|f_n(x)-f(x)|<\varepsilon
\end{align*}
for every $n\ge N$. Since $x\in E$ and $\varepsilon>0$ were arbitrary, $f_n(x)\to f(x)$ for every $x\in E$. Thus $f_n\to f$ pointwise on $E$.
[/step]
[step:Control the exceptional sets by the $L^p$ error]
Assume $1\le p<\infty$, $f_n\in L^p(E,\mathcal{E},\mu)$ for every $n\in\mathbb{N}$, $f\in L^p(E,\mathcal{E},\mu)$, and $\|f_n-f\|_{L^p(E,\mathcal{E},\mu)}\to 0$. Fix $\varepsilon>0$. For each $n\in\mathbb{N}$, define the measurable exceptional set
\begin{align*}
B_n(\varepsilon):=\{x\in E: |f_n(x)-f(x)|>\varepsilon\}.
\end{align*}
On $B_n(\varepsilon)$ we have $\varepsilon^p<|f_n-f|^p$, and therefore monotonicity of the integral gives
\begin{align*}
\varepsilon^p\mu(B_n(\varepsilon))\le \int_{B_n(\varepsilon)} |f_n-f|^p\,d\mu(x)\le \int_E |f_n-f|^p\,d\mu(x).
\end{align*}
Equivalently,
\begin{align*}
\mu(B_n(\varepsilon))\le \varepsilon^{-p}\|f_n-f\|_{L^p(E,\mathcal{E},\mu)}^p.
\end{align*}
The right-hand side tends to $0$ as $n\to\infty$. Hence for every $\varepsilon>0$,
\begin{align*}
\mu(\{x\in E: |f_n(x)-f(x)|>\varepsilon\})\to 0,
\end{align*}
which is exactly convergence in measure.
[guided]
Assume $1\le p<\infty$, $f_n\in L^p(E,\mathcal{E},\mu)$ for every $n\in\mathbb{N}$, $f\in L^p(E,\mathcal{E},\mu)$, and $\|f_n-f\|_{L^p(E,\mathcal{E},\mu)}\to 0$. To prove convergence in measure, we must show that for each fixed threshold $\varepsilon>0$, the measure of the set where the error exceeds $\varepsilon$ tends to $0$.
Fix $\varepsilon>0$. For each $n\in\mathbb{N}$, define
\begin{align*}
B_n(\varepsilon):=\{x\in E: |f_n(x)-f(x)|>\varepsilon\}.
\end{align*}
This set belongs to $\mathcal{E}$ because $f_n-f:E\to\mathbb{R}$ is measurable and $(\varepsilon,\infty)\subset\mathbb{R}$ is Borel after applying the measurable map $x\mapsto |f_n(x)-f(x)|$.
The reason this set is useful is that the $L^p$ integral has a fixed lower bound on it. Indeed, for every $x\in B_n(\varepsilon)$,
\begin{align*}
\varepsilon^p<|f_n(x)-f(x)|^p.
\end{align*}
By monotonicity of the integral over the measurable set $B_n(\varepsilon)$,
\begin{align*}
\varepsilon^p\mu(B_n(\varepsilon))\le \int_{B_n(\varepsilon)} |f_n-f|^p\,d\mu(x).
\end{align*}
Since $B_n(\varepsilon)\subset E$ and $|f_n-f|^p\ge 0$, expanding the domain of integration can only increase the integral:
\begin{align*}
\int_{B_n(\varepsilon)} |f_n-f|^p\,d\mu(x)\le \int_E |f_n-f|^p\,d\mu(x).
\end{align*}
Combining these inequalities gives
\begin{align*}
\mu(B_n(\varepsilon))\le \varepsilon^{-p}\int_E |f_n-f|^p\,d\mu(x)=\varepsilon^{-p}\|f_n-f\|_{L^p(E,\mathcal{E},\mu)}^p.
\end{align*}
The hypothesis $\|f_n-f\|_{L^p(E,\mathcal{E},\mu)}\to 0$ implies that the right-hand side tends to $0$. Therefore
\begin{align*}
\mu(\{x\in E: |f_n(x)-f(x)|>\varepsilon\})\to 0.
\end{align*}
Since this holds for every $\varepsilon>0$, $f_n\to f$ in measure.
[/guided]
[/step]
[step:Convert almost everywhere convergence into shrinking tail sets]
Assume $\mu(E)<\infty$ and $f_n\to f$ $\mu$-almost everywhere on $E$. Fix $\varepsilon>0$. By almost everywhere convergence, there exists a measurable set $Z\in\mathcal{E}$ with $\mu(Z)=0$ such that $f_n(x)\to f(x)$ for every $x\in E\setminus Z$.
For each $n\in\mathbb{N}$, define
\begin{align*}
A_n(\varepsilon):=\bigcup_{k\ge n}\{x\in E: |f_k(x)-f(x)|>\varepsilon\}.
\end{align*}
Each $A_n(\varepsilon)$ is measurable, and the sequence $(A_n(\varepsilon))_{n=1}^{\infty}$ is decreasing because increasing $n$ removes sets from the union. Also $A_1(\varepsilon)\subset E$, so $\mu(A_1(\varepsilon))<\infty$.
We claim that
\begin{align*}
\bigcap_{n=1}^{\infty} A_n(\varepsilon)\subset Z.
\end{align*}
Indeed, if $x\in E\setminus Z$, then $f_k(x)\to f(x)$, so there exists $N_x\in\mathbb{N}$ such that $|f_k(x)-f(x)|\le \varepsilon$ for every $k\ge N_x$. Hence $x\notin A_{N_x}(\varepsilon)$, and therefore $x\notin \bigcap_{n=1}^{\infty}A_n(\varepsilon)$.
By continuity from above for the finite-measure decreasing sequence $(A_n(\varepsilon))_{n=1}^{\infty}$,
\begin{align*}
\mu(A_n(\varepsilon))\to \mu\left(\bigcap_{n=1}^{\infty}A_n(\varepsilon)\right).
\end{align*}
The intersection is contained in $Z$, so its measure is $0$. Therefore $\mu(A_n(\varepsilon))\to 0$. Since
\begin{align*}
\{x\in E: |f_n(x)-f(x)|>\varepsilon\}\subset A_n(\varepsilon),
\end{align*}
we obtain
\begin{align*}
\mu(\{x\in E: |f_n(x)-f(x)|>\varepsilon\})\to 0.
\end{align*}
Because $\varepsilon>0$ was arbitrary, $f_n\to f$ in measure.
[/step]
