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I'm trying to prove the following statements in Folland's book.

Let $(X,\mathcal{M},\mu)$ be a measure space. If $f_n\to f$ in measure and $g_n\to g$ in measure, then $f_n+g_n\to f+g$ in measure and $f_ng_n\to fg$ in measure (in this last case, assume $\mu(X)<\infty$).

The first one is already done, but the second one is giving me trouble.

My attempt: We can take subsequences $f_{n_k}$ and $g_{n_k}$ such that $f_{n_k}\to f$ a.e. and $g_{n_k}\to g$ a.e. In this case, we know that $f_{n_k}g_{n_k}\to fg$ a.e. Note that $\mu(X)<\infty$, so we can use Egoroff's theorem to conclude that $f_{n_k}g_{n_k}\to fg$ almost uniformly. Finally, almost uniform convergence implies convergence in measure, therefore $f_{n_k}g_{n_k}\to fg$ in measure.

The problem is that I just have convergence in measure for some subsequence of $f_ng_n$ and I need convergence in measure for the whole sequence. Also I don't know if using Egoroff's theorem is unnecessary, if this is the case, I would appreciate to see some more elementary prove.

Detail: We say that $f_n\to f$ almost uniformly if for every $\varepsilon > 0$ there is some $E\subset X$ measurable, such that $\mu(E) <\varepsilon$ and $f_n\to f$ uniformly on $E^c$.

Thank you!

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There is something which is sometimes called the subsequence principle. It states that $x_n \to x$ holds if and only if every subsequence $(x_{n_k})_k$ admits a further subsequence $(x_{n_{k_l}})_l$, which converges to $x$. Note that the limit $x$ has to be fixed.

This holds as soon as the notion of convergence is induced by a topology.

This should help you here.

Interesting sidenote: Using this property, one can show that convergence a.e. (of measurable functions on $\Bbb{R}$) is not induced by a topology.

For the proof of the principle, note that "$\Rightarrow$" is trivial (use $(x_{n_k})_k$ itself for the "further subsequence". For "$\Leftarrow$", assume that $x_n \to x$ is false. Then (by definition) there is a neighborhood $U$ of $x$ such that for each $N\in \Bbb{N}$, there is some $n_N \geq N$ with $x_{n_N}\notin U$. Inductively, this allows to construct a subsequence $(x_{n_k})_k$ with $x_{n_k}\notin U$ for all $k$. But by assumption, there is a further subsequence $(x_{n_{k_l}})_l$ which converges to $x$. In particular, for $l$ large enough, $x_{n_{k_l}}\in U$ (because $U$ is an neighborhood of $x$), contradiction.

EDIT: Here is another possible proof: Using the subsequence principle, you can also show the following equivalence on a space of finite measure:

A sequence $(f_n)_n$ of measurable functions converges in measure to $f$ iff every subsequence $(f_{n_k})_k$ has a further subsequence $(f_{n_{k_\ell}})_\ell$ which converges pointwise a.e. to $f$.

Once you have shown this characterization, you can derive your claim as a corollary.

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  • $\begingroup$ Do you have a link with the proof of this principle. I don't want to use results without knowing their proofs. $\endgroup$
    – Integral
    Jan 25 '15 at 16:25
  • $\begingroup$ Don't worry about that anymore, I could help myself. $\endgroup$
    – Integral
    Jan 25 '15 at 17:03
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    $\begingroup$ We now there is a subsequence of $\{f_2, f_4, \ldots\}$ converging a.e. to $f$, suppose this subsequence is $\{f_{10}, f_{20},\ldots\}$, with the index being all multiples of $10$. On the other side, we know there is a subsequence of $\{g_2, g_4,\ldots\}$ converging a.e. to $g$, suppose this subsequence is $\{g_{12}, g_{22},\ldots\}$, with the index being all multiples of $10$ plus $2$. In this view, there is no way in can use the product os this subsequences to argue something, because we need products of the form $f_ng_n$, the indexes must agree. $\endgroup$
    – Integral
    Jan 26 '15 at 14:11
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    $\begingroup$ In your example, consider the sequence $(h_n)_n = \{g_10, g_20, \dots\}$. This sequence still converges in measure to $g$, since it is a subsequence of $(g_n)_n$. Hence, there is a subsequence of $(h_n)_n$, which maybe takes each 3rd entry of $(h_n)_n$ such that $(h_{3n})_n$ converges pointwise a.e. to $g$. But $(h_{3n})_n = (g_{30}, g_{90}, \dots)$. Since $(f_{30}, f_{90}, \dots)$ is also a subsequence of $(f_{10}, f_{20}, \dots)$, which we know converges pointwise a.e., everything is fine. $\endgroup$
    – PhoemueX
    Jan 26 '15 at 14:20
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    $\begingroup$ Now it's 100% done. Thank you very much your patience. I really learned a lot. $\endgroup$
    – Integral
    Jan 26 '15 at 22:49
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I prove PhoemueX's claim as follows.

Let $(X,\mathcal{F},\mu)$ be a measure space with $\mu(X)<\infty$. Let $f_{n}:X\rightarrow\mathbb{R}$ and $f:X\rightarrow\mathbb{R}$ be measurable functions. Then the following are equivalent:

(1) $f_{n}\rightarrow f$ in measure,

(2) Every subsequence of $\{f_{n}\}$ has a subsequence that converges to $f$ a.e.

Proof: To prove $(1)\Rightarrow(2)$: Let $\{f_{n_{k}}\}$ be a subsequence of $\{f_{n}\}$. To simplify notation, we denote $g_{k}=f_{n_{k}}$. Clearly $g_{n}\rightarrow f$ in measure. Choose integers $n_{k}$ inductively such that $n_{1}<n_{2}<\ldots$ and $\mu\left(\left[|g_{n_{k}}-f|\geq\frac{1}{k}\right]\right)\leq\frac{1}{2^{k}}$. To further simplify notation, let $h_{k}=g_{n_{k}}$. We assert that $h_{n}\rightarrow f$ a.e.. Define $A=\{x\in X\mid\lim_{n}h_{n}(x)=f(x)\}$. Observe that $$ A^{c}=\bigcup_{l=1}^{\infty}\bigcap_{N=1}^{\infty}\bigcup_{n=N}^{\infty}\{x\mid|h_{n}(x)-f(x)|\geq\frac{1}{l}\}. $$ We go to show that, for each $l\in\mathbb{N}$, $\mu\left(\bigcap_{N=1}^{\infty}\bigcup_{n=N}^{\infty}\{x\mid|h_{n}(x)-f(x)|\geq\frac{1}{l}\}\right)=0$. For each $N$, denote $B_{N}=\bigcup_{n=N}^{\infty}\{x\mid|h_{n}(x)-f(x)|\geq\frac{1}{l}\}$. For $n\geq l$, we have $\{x\mid|h_{n}(x)-f(x)|\geq\frac{1}{l}\}\subseteq\{x\mid|h_{n}(x)-f(x)\mid\geq\frac{1}{n}\}$, so for any $N\geq l$, $$ \mu\left(B_{N}\right)\leq\sum_{n=N}^{\infty}\mu\left(\{x\mid|h_{n}(x)-f(x)|\geq\frac{1}{l}\}\right)\leq\sum_{n=N}^{\infty}\mu\left(\{x\mid|h_{n}(x)-f(x)|\geq\frac{1}{n}\}\right)\leq\sum_{n=N}^{\infty}\frac{1}{2^{n}}. $$ Observe that $B_{1}\supseteq B_{2}\supseteq\ldots$ and $\mu$ is finite, we have $\mu(\cap_{N}B_{N})=\lim_{N\rightarrow\infty}\mu(B_{N})=0$. Therefore, $\mu(A^{c})=0$ and hence $h_{n}\rightarrow f$ a.e.

To prove $(2)\Rightarrow(1)$: Prove by contradiction. Suppose the contrary that $f_{n}\not\rightarrow f$ in measure. Then there exists $\delta_{1}>0$ such that $\mu\left(\left[|f_{n}-f|\geq\delta_{1}\right]\right)\not\rightarrow0$. There further exist $\delta_{2}>0$ and a subsequence $(n_{k})$ such that for each $k$, $\mu\left(\left[|f_{n_{k}}-f|\geq\delta_{1}\right]\right)\geq\delta_{2}.$ Define $\delta=\min(\delta_{1},\delta_{2})$. In particular, we have $\mu\left(\left[|f_{n_{k}}-f|\geq\delta\right]\right)\geq\delta$. To further simplify notation, denote $g_{k}=f_{n_{k}}$. Then $(g_{n})$ is a subsequence of $(f_{n})$ and $\mu\left(\left[|g_{n}-f|\geq\delta\right]\right)\geq\delta$ for each $n$. By assumption, $(g_{n})$ has a subsequence $(g_{n_{k}})$ that converges to $f$ a.e.. Define $D=\{x\mid g_{n_{k}}(x)\rightarrow f(x)\}$, then $$ D^{c}=\bigcup_{l=1}^{\infty}\bigcap_{N=1}^{\infty}\bigcup_{k=N}^{\infty}\{x\mid|g_{n_{k}}(x)-f(x)|\geq\frac{1}{l}\}. $$ Choose $l$ such that $\frac{1}{l}<\delta$. Note that $\mu(D^{c})=0$, and in particular, $\mu\left(\bigcap_{N=1}^{\infty}\bigcup_{k=N}^{\infty}\{x\mid|g_{n_{k}}(x)-f(x)|\geq\frac{1}{l}\}\right)=0.$ Note that $\mu$ is finite and $\left(\bigcup_{k=N}^{\infty}\{x\mid|g_{n_{k}}(x)-f(x)|\geq\frac{1}{l}\}\right)_{N}$ is a decreasing sequence of measurable sets. Therefore $$ \lim_{N\rightarrow\infty}\mu\left(\bigcup_{k=N}^{\infty}\{x\mid|g_{n_{k}}(x)-f(x)|\geq\frac{1}{l}\}\right)=\mu\left(\bigcap_{N=1}^{\infty}\bigcup_{k=N}^{\infty}\{x\mid|g_{n_{k}}(x)-f(x)|\geq\frac{1}{l}\}\right)=0. $$ Choose $N$ such that $\mu\left(\bigcup_{k=N}^{\infty}\{x\mid|g_{n_{k}}(x)-f(x)|\geq\frac{1}{l}\}\right)<\delta$. Observe that for each $k\geq N$, $\{x\mid|g_{n_{k}}(x)-f(x)|\geq\frac{1}{l}\}\supseteq\{x\mid|g_{n_{k}}(x)-f(x)|\geq\delta\}$, so $$ \mu\left(\bigcup_{k=N}^{\infty}\{x\mid|g_{n_{k}}(x)-f(x)|\geq\frac{1}{l}\}\right)\geq\mu\left(\{x\mid|g_{n_{N}}(x)-f(x)|\geq\delta\}\right)\geq\delta $$ which is a contradiction.

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By using the claim, we can prove that $f_{n}\rightarrow f$ and $g_{n}\rightarrow g$ in measure $\Rightarrow$ $f_{n}g_{n}\rightarrow fg$ in measure immediately.

For, let $\left(f_{n_{k}}g_{n_{k}}\right)_{k}$ be an arbitrary subsequence of $(f_{n}g_{n})_{n}$. $f_{n}\rightarrow f$ in measure implies that for the subsequence $(f_{n_{k}})_{k}$, there exists a subsequence $(f_{n_{k_{l}}})_{l}$ such that $f_{n_{k_{l}}}\rightarrow f$ a.e.. $g_{n}\rightarrow g$ in measure implies that for the subsequence $(g_{n_{k_{l}}})_{l}$, there exists a further subsequence $(g_{n_{k_{l_{p}}}})_{p}$ such that $g_{n_{k_{l_{p}}}}\rightarrow g$ a.e.. Now, we have $f_{n_{k_{l_{p}}}}g_{n_{k_{l_{p}}}}\rightarrow fg$ a.e.. By the claim, we have $f_{n}g_{n}\rightarrow fg$ in measure.

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This is the alternative method mentioned by yi li's answer on this webpage.

Define functions $s_n,d_n,s,d$ on $X$ by\begin{align*} s_n(x) & = (f_n(x) + g_n(x))/4,\\ d_n(x) & = (f_n(x) - g_n(x))/4,\\ s(x) & = (f(x) + g(x))/4,\\ d(x) & = (f(x) - g(x))/4\end{align*} all of which are finite a.e. on $X$ and measurable. Moreover, $\{s_n\}\to s$ and $\{d_n\}\to d$ in measure. Then\begin{align*} &\lim_{n\to\infty}\mu\{x\mid |f_n(x)g_n(x) - f(x)g(x)| >\eta\}\\ &\quad =\lim_{n\to\infty}\mu\{x\mid |s_n^2(x) - d_n^2(x) - s^2(x) + d^2(x)| >\eta\}\\ &\quad\le\lim_{n\to\infty}\mu\{x\mid |s_n^2(x) - s^2(x)| >\eta/2\}\\ &\quad+\lim_{n\to\infty}\mu\{x\mid |d_n^2(x) - d^2(x)| >\eta/2\}\\ &\quad = 0 + 0.\end{align*} Thus, $\{f_n g_n\}\to fg$ in measure.

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An alternative method is based on the result here product case

And the rest of proof is based on the first result

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