Let $E\subset\mathbb{R}^n$ be a measurable set with finite Lebesgue measure, $\{f_n\}_{n∈N}$ be a sequence of measurable functions $f_n : E → \mathbb{R}$, bounded in $L^p(E)$ for $p>1$, and pointwise converging to $f : E → \mathbb{R}$ almost everywhere.
Since the sequence converge almost everywhere, consider the set $N_0$ of measure zero where the sequence does not converge, and $E\setminus N_0$ where the sequence converge. By Egorov theorem the sequence converges to $f$ uniformly on $E\setminus N_0$.
Study the convergence in $L^q(E)$ for $q∈(1,p)$: \begin{align*} ||f_n-f||_{L^q(E)}&\le||f_n-f||_{L^q(N_0)}+||f_n-f||_{L^q(E\setminus N_0)} \\ &\le ||f_n||_{L^q(N_0)}+||f||_{L^q(N_0)}+||f_n-f||_{L^q(E\setminus N_0)} \end{align*} I'd say that the first two norms in the rhs are $0$ since they are integrals on a set of measure $0$, but wouldn't this mean that $f_n$ converge to $f$ on $N_0$? But it is a contraddiction because we defined it as the set where the sequence does not converge, isn'it?
Moreover how to compute or bound the last norm on the rhs?
First, we note that Fatou's Lemma implies that $f \in L^p(E)$. Now, we apply Egorov and obtain that for every $\varepsilon > 0$, there exists a measurable set $M_\varepsilon$ with $\mu(M_\varepsilon) \le \varepsilon$ such that $f_n$ converges uniformly to $f$ on $E \setminus M_\varepsilon$. Now, we use the triangle inequality to get $$\|f_n - f\|_{L^q(E)} \le \|f_n - f\|_{L^q(M_\varepsilon)} + \|f_n-f\|_{L^q(E\setminus M_\varepsilon)} \le \|f_n - f\|_{L^q(M_\varepsilon)} + \|f_n-f\|_{L^\infty(E\setminus M_\varepsilon)} \, \mu(E)^{1/q}.$$ In the first addend, we can use Hölders inequality (with $1/q = 1/p + 1/r$ for some $r \in (1,\infty)$) to obtain $$\|f_n - f\|_{L^q(M_\varepsilon)} \le \|1\|_{L^r(M_\varepsilon)} \, \|f_n - f\|_{L^p(M_\varepsilon)} \le \varepsilon^{1/r} \, (\|f_n\|_{L^p(M_\varepsilon)}+\|f\|_{L^p(M_\varepsilon)} \le C \, \varepsilon^{1/r}.$$ Thus, $$\|f_n - f\|_{L^q(E)} \le C \, \varepsilon^{1/r} + \|f_n-f\|_{L^\infty(E\setminus M_\varepsilon)} \, \mu(E)^{1/q}.$$ Hence, we can choose $N$ (depending on $\varepsilon$) large enough such that $$\|f_n - f\|_{L^q(E)} \le 2 \, C \, \varepsilon^{1/r} \qquad\forall n \ge N.$$
Hence, $f_n \to f$ in $L^q(E)$. Note that this argument also works for $q = 1$.