Let $E_{i,j} = \{x\in E: |f_j(x) - f(x)|< 1/i\}.$ Since $f_n\to f$ almost everywhere, there is a set $S$ with $\mu(S)=0$ such that, given $i\in \N$ and $x\in E-S$ there is $m\in \N$ such that $j>m$ implies $|f_j(x)-f(x)|<1/i$ This can be expressed by $$E-S\subset \cup_{m\in \N} \cap_{j>m} E_{i,j},$$ or, in other words, $$\cap_{m\in \N}\cup_{j>m} (E-E_{i,j})\subset S.$$ Since $\{\cup_{j>m} (E-E_{i,j})\}_{m\in \N}$ is a decreasing nested sequence of sets, each of which has finite measure, and such that its intersection has measure $0$ by continuity from above we know that $$\mu(\cup_{j>m}(E-E_{i,j}))\xrightarrow[m\to \infty]{} 0.$$ Therefore, for each $i\in \N$ we can choose $m_i$ such that $$\mu(\cup_{j>m_i}(E-E_{i,j})) < \frac{\delta}{2^i}.$$ Let $$E_\delta = \cup_{i\in \N}\cup_{j>m_i}(E-E_{i,j}).$$ Then $$\mu(E_\delta)\leq \sum_{i=1}^\infty \mu(\cup_{j>m_i}(E-E_{i,j})) < \sum_{i=1}^\infty \frac{\delta}{2^i} = \delta.$$ We claim that $f_n\to f$ uniformly on $E-E_\delta$ In fact, given $\varepsilon>0$ choose $n$ such that $1/n<\varepsilon$ If $x\in E-E_\delta$ we have $$x\in\cap_{i\in \N}\cap_{j>m_i}E_{i,j},$$ which in particular implies that, if $j>m_n$ $x\in E_{n,j}$ that is, $|f_j(x) - f(x)|< 1/n < \varepsilon$ Hence, for each $\varepsilon>0$ there is $N$ (which is given by $m_n$ above) such that $j>N$ implies $|f_j(x)-f(x)|<\varepsilon$ for each $x\in E-E_\delta$ as required. This completes the proof.