Vrftsjtryk

Let $Esubsetmathbb{R}^n$ 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 $Esetminus N_0$ where the sequence converge. By Egorov theorem the sequence converges to $f$ uniformly on $Esetminus N_0$.

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(Esetminus N_0)} \ &le ||f_n||_{L^q(N_0)}+||f||_{L^q(N_0)}+||f_n-f||_{L^q(Esetminus 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
$$|f_n - f|_{L^q(E)}
le |f_n - f|_{L^q(M_varepsilon)} + |f_n-f|_{L^q(Esetminus M_varepsilon)}
le |f_n - f|_{L^q(M_varepsilon)} + |f_n-f|_{L^infty(Esetminus M_varepsilon)} , mu(E)^{1/q}.$$
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,
$$|f_n - f|_{L^q(E)}
le
C , varepsilon^{1/r}
+ |f_n-f|_{L^infty(Esetminus M_varepsilon)} , mu(E)^{1/q}.$$
Hence, we can choose $N$ (depending on $varepsilon$) large enough such that
$$|f_n - f|_{L^q(E)}
le
2 , C , varepsilon^{1/r}
qquadforall n ge N.$$

Hence, $f_n to f$ in $L^q(E)$. Note that this argument also works for $q = 1$.