2
$\begingroup$

Let $(X, M, m)$ be a finite measurable space and $f, (f_n)_n, g$ measurable functions. Then the following hold:

Proposition 1: If $(f_n)_n$ converges in measure m to both $f$ and $g$, then $f = g$ almost everywhere.

Proposition 2: $(f_n)_n$ converges in measure to f iff for every subsequence $(f_{k_n})_n$ of $(f_n)_n$, there exists a further subsequence $(f_{k_{l_n}})_n$ such that $(f_{k_{l_n}})_n$ converges to $f$ m-almost everywhere.

I am interested in the proof of proposition 1 as a direct collorary of proposition 2. It was stated in class that the proof should follow naturally but I'm struggling to see this.

Please note that I am aware of the 'common' proof of proposition 1 as found e.g. here.

$\endgroup$
  • $\begingroup$ So, you are struggling to see why almost everywhere convergence implies almost everywhere uniqueness? $\endgroup$ – Will M. Jan 15 '18 at 20:05
2
$\begingroup$

It is not hard to see that, if $(f_{n})$ converges to $f$ (or $g$) in measure, then all the subsequences $(f_{n_{k}})$ converges to $f$ (or $g$) in measure as well.

Now given $(f_{n})$ converges to $f$ in measure, then by Proposition 2 there exists some subsequence $(f_{n_{k}})$ converges to $f$ $m$-almost everywhere, say, $f_{n_{k}}(x)\rightarrow f(x)$ for all $x\in X-N_{1}$, $m(N_{1})=0$.

Now for this subsequence $(f_{n_{k}})$, it also converges to $g$ in measure, again by Proposition 2, then there exists some subsequence $(f_{n_{k_{l}}})$ such that $f_{n_{k_{l}}}(x)\rightarrow g(x)$ for all $x\in X-N_{2}$, $m(N_{2})=0$.

Now $m(N_{1}\cup N_{2})=0$ and for all $x\in X-(N_{1}\cup N_{2})$, we have both $f_{n_{k_{l}}}(x)\rightarrow f(x)$ and $f_{n_{k_{l}}}(x)\rightarrow g(x)$, so $f(x)=g(x)$ for all such $x$, hence $f=g$ $m$-almost everywhere.

| cite | improve this answer | |
$\endgroup$
2
$\begingroup$

The sequence $(f_n)_{n \in \mathbb{N}}$ is trivially a subsequence of $(f_n)_{n \in \mathbb{N}}$, and therefore it follows from Proposition 2 that there exists a subsequence $(f_{n_k})_{k}$ such that $f_{n_k} \to f$ $m$-almost everywhere. Applying Proposition 2 once more we find that $f_n \to g$ in measure implies that there exists a subsequence $(f_{n_{k_{\ell}}})_{\ell}$ of $(f_{n_k})_{k}$ which converges $m$-almost everyhwere to $g$. However, $(f_{n_{k_{\ell}}})_{\ell}$ is a subsequence of $(f_{n_k})_k$, and therefore we also have $f_{n_{k_{\ell}}} \to f$ $m$-almost everywhere. Hence, by the uniqueness of pointwise limits, we conclude $f=g$ almost everywhere.

| cite | improve this answer | |
$\endgroup$

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.