[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.[/step]

[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: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]