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I have an issue with the solution to the following problem. I now want to prove that the Lebesgue Dominated Convergence Theorem still works when the condition $\{ f_{n} \}$ converges to $f$ a.e. is replaced by $\{ f_{n} \}$ converges to $f$ in measure.

Relax, I know a lot of people have asked this question already. But what I do not understand is that the solutions I found online insist to work with a sub-subsequence instead of just a subsequence. For instance, the following is an expressed answer I summarise from someone:

We can take a subsequence $\{f_{n_{k}} \}$ converges to $f$ in measure. Then, by a proved proposition, there is a sub-subsequence $\{f_{n_{k_{l}}} \}$ converges to $f$ a.e.. Then it reduces to the usual LDCT conditions and we can have the conclusion.

But how about I simplify a little bit? Can't I just take a subsequence $\{f_{n_{k}} \}$ which converges to $f$ a.e. and have the conclusion? I fail to see the need to take a subsequence that converges to $f$ in measure first. Is that really necessary?

BM Yoon
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  • Would it be that my way just proves LDCT could only apply to one subsequence of ${f_{n}}$, but using a sub-subsequence would allow me to show that LDCT could apply to any subsequence and hence, ${f_{n}}$? – BM Yoon Oct 01 '18 at 15:44
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    You seem to get the point. The reason for considering further subsequences is that we want to facilitate the following equivalence, which is quite a standard machinery when you are only allowed to extract a subsequence with desired property: For a sequence $(x_n)$ and a point $x$ in a metric space, the followings are equivalent. (1) $(x_n)$ converges to $x$. (2) Every subsequence of $(x_n)$ has a further subsequence that converges to $x$. – Sangchul Lee Oct 01 '18 at 16:01

1 Answers1

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If $f_n$ converges to $f$ in measure then there is a subsequence $f_{n_k}$ that converges to $f$ almost everywhere, and so we could apply Lebsegue dominated convergence to that subsequence, which is not good enough.

So the proof does seem to need a sub-sub-sequence as follows:

Suppose $|f_n(x)|\leq g(x)$ for all $x \in X$, $\int g < \infty$, and $f_n$ converges to $f$ in measure. Then $$ \lim_{n\rightarrow\infty} \int f_n = \int f $$

Proof: Suppose not (we reach a contradiction). Then there is an $\epsilon>0$ and a subsequence $f_{n_k}$ such that $$\left|\int f_{n_k} - \int f\right| \geq \epsilon \quad \forall k \in \{1, 2, 3, ...\} \quad \mbox{ (Eq. 1)}$$ Now $f_{n_k}$ converges to $f$ in measure so there is a sub-subsequence $f_{n_{k[m]}}$ that converges to $f$ almost everywhere. So the usual Lebsegue dominated convergence theorem applies to that sub-subsequence to ensure $$ \lim_{m\rightarrow\infty}\int f_{n_{k[m]}} = \int f$$ which contradicts (Eq. 1). $\Box$

Michael
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