Convergence of a sequence in two different $L_p$ spaces If I have a sequence $\{f_n\}_{n\in \mathbb{N}} \subset L_p\cap L_q$ that is Cauchy under both norms,  I am wondering if $f_n \to f$ in $L_p$ implies that $f_n\to f$ in $L_q$. 
I have been working on this but I can't seem to figure out a definite proof.
Thank you!
PS: $1\leq p<q <\infty$
 A: Yes, if the sequence is a Cauchy sequence in both norms, it converges to the same function in both norms. That follows from the fact that a sequence converging in $L^r,\; 1 \leqslant r \leqslant \infty$, has a subsequence that converges almost everywhere to the limit function.
Since the $L^r$ spaces are complete, the condition that $(f_n)$ be a Cauchy sequence in both norms asserts the existence of limits in both spaces,
$$f_n \xrightarrow[n\to\infty]{L^p} f^{(p)};\qquad f_n \xrightarrow[n\to\infty]{L^q} f^{(q)}.$$
We can now extract a subsequence $(f_{n_k})$ that converges pointwise almost everywhere to the $L^p$-limit $f^{(p)}$. That subsequence is still a Cauchy sequence in both norms, and converges to $f^{(p)}$ resp. $f^{(q)}$ in the two spaces.
Since $f_{n_k} \xrightarrow[k\to\infty]{L^q} f^{(q)}$, we can extract a subsequence $(f_{n_{k_m}})$ of $(f_{n_k})$ that converges pointwise almost everywhere to $f^{(q)}$.
Then the subsequence $(f_{n_{k_m}})$ of $(f_n)$ converges pointwise almost everywhere to $f^{(p)}$, since it is a subsequence of $(f_{n_k})$, and it converges pointwise almost everywhere to $f^{(q)}$. Thus we have $f^{(p)} = f^{(q)}$ almost everywhere, and the $L^p$-limit is the same as the $L^q$-limit (as a class of measurable functions modulo the equivalence relation of being equal almost everywhere).
