Convergence in measure implies pointwise convergence?

Question

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?

Answer 1

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.

Answer 2

I see what you intended now by that web link. Suppose $f_n$ converges to $f$ in measure, and there is a $g$ such that $|f_n|\leq g$ for all $n$, and $\int g < \infty$.

I think you mean this: Your original sequence is $\{\int f_n\}_{n=1}^{\infty}$. Consider an infinite subsequence of this with indices in $\mathcal{N}$. We want to show that there exists a convergent subsequence $\int f_{n_k}$, with $n_k \in \mathcal{N}$ for all $k$ (and where $n_k < n_{k+1}$ for all $k$) that satisfies: $$ \int f_{n_k} \rightarrow \int f $$ If this is true, then by that web link, we also know that $\int f_n \rightarrow \int f$.

The good thing is that if a sequence of functions converges to $f$ in measure, then there is indeed a subsequence $f_{n_k}$ that converges to $f$ pointwise almost everywhere. So then we can invoke the usual Lebesgue theorem to ensure $\int f_{n_k} \rightarrow \int f$.

I was indeed confused by your original use of "converging pointwise" when it should really be "converging pointwise almost everywhere," as another person commented.

Proving that claim: Suppose $h_n$ converges to $h$ in measure. Then there is a subsequence $h_{n_k}$ that converges to $h$ pointwise almost everywhere.

I think this can be proven in the same way as the probability fact that if random variables $X_n$ converge to $X$ in probability, a subsequence converges with prob 1. The main step is to define a subsequence of functions $h_{n_k}$ such that for each $k$: $$\mu(\{x \mbox{ such that } |h_{n_k}(x)-h(x)|>1/k\}) \leq 1/k^2 $$

Answer 3

1. There is nothing wrong with the statement if $f_n\to 0$ in measure, then so does every subsequence $f_{n_k}$. It is wrong, however, to state that if a subsequence $\{f_{n_k}\}$ converges pointwise, then so must $\{f_n\}$.

2. No.

3. (and 4) See above.