Convergence in measure implies pointwise convergence?

Solution 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.

Solution 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 $k$: $$\mu(\{x \mbox{ such that } |h_{n_k}(x)-h(x)|>1/k\}) \leq 1/k^2 $$