If a sequence $(u_n)$ in $L^p \cap L^q$ converges to $u$ in $L^p$ and to $v$ in $L^q$, then $u=v$ a.e.

Question

Is it true that if a sequence $(u_n)$ in $L^p \cap L^q$ converges to $u$ in $L^p$ and to $v$ in $L^q$, then $u=v$ almost everywhere?

Since $u_{k}\to u$ in $L^p$, there exists a subsequence $(u_{k_i})$ such that $u_{k_i}\to u$ almost everywhere.

Similarly, since $u_{k}\to v$ in $L^q$, there exists a subsequence $(u_{l_i})$ such that $u_{l_i}\to v$ almost everywhere.

But I think this is not enough to conclude that $u=v$ almost everywhere. Any help would be greatly appreciated!

Answer 1

This is also very similar to your previous two questions. (at least, the two that I saw)

Let $U = \{x: u \neq v\} = \bigcup_{n=1}^\infty\{ x: |u(x) - v(x) | > 1/n \}.$ Note triangle inequality implies, for any $n,k$: $$ |\{ x: |u(x) - v(x) | > 1/n \}| \le |\{ x: |u(x) - u_k(x) | > 1/2n \}| + |\{ x: |u_k(x) - v(x) | > 1/2n \}|$$ Taking $k$ large we see by Chebyshev/Markov (not sure which name it is) $$ |\{ x: |u(x) - u_k(x) | > 1/2n \}| \le 2^pn^{p} \| u-u_k\|^p_{L^p} \le \epsilon, $$ so $\{ x: |u(x) - v(x) | > 1/n \}$ has arbitrarily small measure, i.e. its a null set; thus by union bound $$ |U| \le \sum_1^\infty |\{ x: |u(x) - v(x) | > 1/n \}| = \sum_1^\infty 0 = 0.$$

The proof of Chebyshev/Markov is so simple that I don't remember the statement and instead remember the one-line proof- $$\int_X |f|^p d\mu \ge \int_{|f|>1/n} |f|^p d\mu \ge n^{-p} \int_{|f|>1/n} d\mu = n^{-p} \mu(|f|>1/n).$$

Answer 2

The problem you're having is similar to one encountered when using Boltzano-Weierstrass unwisely. Since $\{u_{k_i}\}$ is a sub-sequence of $\{ u_k\}$, $u_{k_i}$ also converges to $v$ in $L^q$. Now you can a take a further sub-sequence, $\{ u_{\ell_i} \}$, converging almost everywhere to $v$.

As a subsequence of a sequence already converging to $u$ almost everywhere, it would converge both to $u$ almost everywhere and to $v$ almost everywhere.