Norm of a function on a set of measure $0$ and convergence of a sequence on a set of finite measure 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?
 A: 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 $q = 1$.
