In showing that we can replace pointwise convergence with convergence in measure in the Lebesgue Dominated Convergence Theorem, I made the following claim:
1.) $f_n\to f$ in measure $\,\,\Longrightarrow\,\,$ every $f_{n_k}\to f$ in measure $\,\,\Longrightarrow\,\,$ some $f_{n_{k_j}}\to f$ pointwise.
2.) Then since $\{f_n\}$ is a sequence such that every subsequence $\{f_{n_k}\}$ has a further subsubsequence $\{f_{n_{k_j}}\}$ that converges pointwise to $f$, $f_n$ converges pointwise to $f$ as well.
But this seems to prove that convergence in measure implies pointwise convergence, which we know to be false. Consider this example:
1.) Let $\{I_n\}_{n=1}^\infty=\{[0,1], [0,1/2], [1/2,1], [0,1/3], [1/3,2/3], [2/3,1], [0,1/4],\ldots\}$.
2.) Let $f_n(x)=\chi_{I_n}(x)$ for all $x\in[0,1]$. According to my text, $f_n\to 0$ in measure but there exists no $x \in [0,1]$ such that $f_n\to 0$ pointwise.
QUESTIONS: The only error in the logic of my original proof seems to be assuming that $f_n\to f$ in measure $\Longrightarrow$ every subsequence $f_{n_k}\to f$ in measure.”
1.) Is the flaw in my proof somewhere else?
2.) Does the sequence of functions in the counterexample have some subsequence that does not converge in measure to 0?
3.) If yes, what is it?
4.) If no, can we create a different sequence that converges measure but has some subsequence that does not converge in measure?
Best Answer
The issue is that you are not taking into consideration that convergence in measure guarantees a subsequence which converges pointwise almost everywhere to $f$ - not everywhere pointwise convergence. Everywhere pointwise convergence is necessary to guarantee that $f_n \to f$ pointwise if every subsequence has a further subsequence which converges to $f$ pointwise.
With that being said, this is an argument that may be used to show that convergence in measure (with a dominating function) gives rise to the dominated convergence theorem. The difference is that convergence in $L^1$ acts on equivalence classes of almost everywhere equal functions.