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?