# Suppose $\{f_k\}$ is a sequence of $M$-measurable functions on $X$. Let $p_1$ and $p_2\in [1,\infty)$, and suppose $f_k\in L^{p_1}\cap L^{p_2}$.

Suppose $\{f_k\}$ is a sequence of $M$-measurable functions on $X$. Let $p_1$ and $p_2\in [1,\infty)$, and suppose $f_k\in L^{p_1}\cap L^{p_2}$. Also suppose there exist $g\in L^{p_1}$ and $h\in L^{p_2}$ such that $f_k\to g$ in $L^{p_1}$, $f_k\to h$ in $L^{p_2}$. Prove that $g=h$ ($\mu$ almost everywhere)

So we know $||f_k-g||_{p_1}\to 0$ and $||f_k-h||_{p_2}\to 0$ as $k\to\infty$. That means $\lim_{k\to\infty}\int |f_k-g|^{p_1}d\mu= 0$ and $\lim_{k\to\infty}\int |f_k-h|^{p_2}d\mu= 0$.

Suppose it's valid to bring the limit inside the integral, and $\lim f_k$ exists (call it $f$). Then $\int|f-g|^{p_1}d\mu=\int|f-h|^{p_2}d\mu=0$. That means $f=g=h$ almost everywhere. But how do I justify bringing the limit inside and $\lim f_k$ exists? Can I apply dominated convergence theorem? Is $|f_k-g|$ bounded? Thanks.

Convergence in $L^p$ spaces implies convergence in measure. Then $$f_k\to g\quad\text{and}\quad f_k\to f$$ in measure. By the uniqueness of the limit of sequences of functions which converges in measure, we can conclude $f = g$ almos everywhere.

Note: Uniqueness of the limit of sequences which converges in measure is a consequence of the Hausdorffness of the topology of "convergence in measure". In general, if a space is Hausdorff then the limits of sequences are unique.

Alternatively:

Since $f_k\to g$ in $L^{p_1}$, there exist a subsequence $(f_{k_j})_j$ such that $f_{k_j}\to g$ pointwise a.e.

Since $f_k\to h$ in $L^{p_2}$, its subsequence $(f_{k_j})_j$ must converge to $h$ as well in $L^{p_2}$.

Since $f_{k_j}\to h$ in $L^{p_2}$, there exist a subsequence $(f_{k_{j_i}})_i$ which converges to $h$ pointwise a.e.

Since $(f_{k_{j_i}})_i$ is a subsequence of $(f_{k_j})_j$ and $f_{k_j}\to g$ pointwise a.e., it follows that $f_{k_{j_i}}\to g$ pointwise a.e.

So, in the set of pointwise convergence of $(f_{k_{j_i}})_i$ we have $g = h$. But this set is the whole space without a set of measure zero, i.e. $g=h$ a.e.

• Can this be proved without using Hausdorffness? We haven't learned it yet. Thanks. – Christmas Bunny Nov 2 '13 at 22:56
• @YifengXu Yes, click on "uniqueness" in my answer. – leo Nov 2 '13 at 22:59
• @YifengXu is there something unclear? – leo Nov 3 '13 at 0:30
• Why does converence in $L^p$ spaces imply convergence in measure? We haven't talked about convergence in measure yet. Thank you! – Christmas Bunny Nov 3 '13 at 0:40
• @YifengXu edited – leo Nov 3 '13 at 4:07

OK, this is a simpler way to say it. If $f_k \to g$ in $L^{p_1}$, then there exists a subsequence $f_{k_n} \to g$ a.e.

$f_k \to g$ in measure, and $f_k \to h$ in measure. And the topology of "convergence in measure" is Hausdorff. http://en.wikipedia.org/wiki/Convergence_in_measure

To show the latter, note that if $f_k \to g$ in measure, then there is a subsequence $f_{k_n}$ that converges a.e. to $g$.

• Can this be proved without resorting to Hausdorff? We haven't learned it yet. Thanks. – Christmas Bunny Nov 2 '13 at 22:56