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.


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.

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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.