Suppose $\{f_n\}$ is a sequence of functions in $L_1$. Show that $\{f_n\}$ is a Cauchy sequence in mean if and only if

$\int_E f_n \, d\mu=x_n$ is a Cauchy sequence for real numbers for every measurable set $E,$ and $\{f_n\}$ is a Cauchy sequence in measure.

I am able to prove the first part:

($\Rightarrow$) Since Cauchy sequence in $p$th mean implies Cauchy sequence in measure, we only need to prove that the sequence $\{x_n\}$ is a Cauchy sequence (We can assume that $E=\Omega$ is the entire space), but that is precisely the definition of being a Cauchy sequence in $L_1$, so we are done

I'm having trouble to proving the second part, since I cannot estimate $\int_E |f_n-f_m| \, d\mu$ using $|x_n-x_m|=|\int_E( f_n-f_m ) \, d\mu|$. I was tying to use one Theorem, too:

Theorem: Suppose $\{f_n\}$ is a sequence of functions is $L_p$ and let

$\nu_n(E)$ = $\int_E |f_n|^p \, d\mu$. Then $\{f_n\}$ is a Cauchy sequence in pth mean if an only if $\{f_n\}$ is a Cauchy sequence in measure and the family $\{\nu_n\}$ is equicontinuous at $\varnothing$.

Since $\nu_n(\Omega)<\infty$, then we only need to show that the family is absolutely continuous with respect to $\mu$, and is precisely here where I'm stuck.

  • $\begingroup$ What is your definition of "absolutely continuous with respect to $\mu$?" Because it is already the case. $\endgroup$ – Guy Fsone Nov 20 '17 at 13:45
  • $\begingroup$ Given $\epsilon>0$, there exists $\delta>0$ such that $\mu(E)<\delta$ implies $\nu_{n}(E)<\epsilon$. I dont see why the family must be absolutely continuous w.r.t $\mu$ $\endgroup$ – alexp9 Nov 20 '17 at 14:13
  • $\begingroup$ it is because (By Radon-Nykodim) $d\nu_n =|f_n|^pd\mu$ this means that $d\nu_n $ is absolute continuous wrt $d\mu$ $\endgroup$ – Guy Fsone Nov 20 '17 at 14:20
  • $\begingroup$ Every function is absolutely continuous wrt to $\mu$. I agree with that. But that is not the same as the whole family being absolutely continuous. Also, note that in our case we only need p=1. Idk if I'm missunderstanding something. Thank you for the help. $\endgroup$ – alexp9 Nov 20 '17 at 14:34
  • $\begingroup$ In my definition, I forgot to put "for all n" just at the end. $\endgroup$ – alexp9 Nov 20 '17 at 14:45

What you are trying to prove is true but the proof is not that easy. It's actually a theorem, called the Vitali-Hahn-Saks theorem.

Let $\left( \Omega,\mathfrak{M},\mu\right) $ be the measure space with $\mu$ finite. By identifying sets that differ by a set of $\mu$ measure zero we can regard $\mathfrak{M}$ as a closed subset $\mathcal{C}$ of $L_{1}$ through the mapping $$ E\in\mathfrak{M}\mapsto\chi_{E}. $$

Fix $\varepsilon>0$ and for $k\in\mathbb{N}$ define the sets $$ \mathcal{C}_{k}:=\left\{ \chi_{E}:\,E\in\mathfrak{M},\,\sup _{n,\,l\geq k}\left\vert \int_{E}f_{n}\,d\mu-\int_{E}f_{l}\,d\mu\right\vert \leq\varepsilon\right\} . $$ We claim that the sets $\mathcal{C}_{k}$ are closed in $L_{1}$. Indeed, if $\left\{ \chi_{E_{j}}\right\} \subset\mathcal{C}_{k}$ converges in $L_{1}$ to some function $f$, then, by extracting a subsequence (not relabeled), if necessary, we may assume that $\chi_{E_{j}}\left( x\right) \rightarrow f(x)$ for $\mu$ a.e. $x\in X$. Since $\chi_{E_{j}}(x)$ takes only values $0$ and $1$, it follows that $f(x)\in\{0,1\}$ for for $\mu$ a.e. $x\in X$. Hence, $f=\chi_{E_{\infty}}$ for some measurable set $E_{\infty}$. Since $\chi_{E_{j}}(x)\leq1$ for every $x\in\Omega$ and all $j$ and $\mu$ is finite, we can apply the Lebesgue dominated convergence theorem to get that for each $n\in\mathbb{N}$, $$ \lim_{j\rightarrow\infty}\int_{E_{j}}f_{n}\,d\mu=\int_{E_{\infty}}f_{n}\,d\mu. $$ Since $\left\{ \chi_{E_{j}}\right\} \subset\mathcal{C}_{k}$ for any fixed $n$,$\,l\geq k$\ and for all $j\in\mathbb{N}$, we have $$ \left\vert \int_{E_{j}}f_{n}\,d\mu-\int_{E_{j}}f_{l}\,d\mu\right\vert \leq\varepsilon, $$ and so letting $j\rightarrow\infty$ we obtain that $$ \left\vert \int_{E_{\infty}}f_{n}\,d\mu-\int_{E_{\infty}}f_{l}\,d\mu \right\vert \leq\varepsilon, $$ which shows that $f\in\mathcal{C}_{k}$. Hence, $\mathcal{C}_{k}$ is closed\ in $L_{1}$.

By the completeness of the reals $\lim_{n\rightarrow\infty}\int_{E}f_{n} \,d\mu$ exists in $\mathbb{R}$ for all $E\in\mathfrak{M}$, which implies that $$ \mathcal{C}=\bigcup_{k=1}\mathcal{C}_{k}. $$ Applying the Baire category theorem to the complete metric space $\mathcal{C} $, at least one of the sets\ $\mathcal{F}_{k}$ has nonempty interior. Hence there exist $\delta_{1}>0$, $k\in\mathbb{N}$, and $\chi_{E_{0}}\in \mathcal{C}_{k}$\ such that if $\chi_{E}\in\mathcal{C}$ and if \begin{equation} \int_{\Omega}\left\vert \chi_{E}-\chi_{E_{0}}\right\vert \,d\mu<\delta _{1},\label{lp vhs1} \end{equation} then $\chi_{E}\in\mathcal{C}_{k}$, that is, \begin{equation} \sup_{n,\,l\geq k}\left\vert \int_{E}f_{n}\,d\mu-\int_{E}f_{l}\,d\mu \right\vert \leq\varepsilon\text{.}\label{lp vhs2} \end{equation} Since each single function $f_{n}$ is integrable, we may find $0<\delta <\delta_{1}$\ such that \begin{equation} \int_{E}\left\vert f_{n}\right\vert \,d\mu\leq\varepsilon\label{lp vhas3} \end{equation} for all $1\leq n\leq k$\ and for every measurable set $E$ with $\mu\left( E\right) \leq\delta$.

Consider a measurable set with $\mu\left( E\right) \leq\delta$. Then we can write $ E=\left( E\cup E_{0}\right) \setminus\left( E_{0}\setminus E\right)$, with $$ \int_{\Omega}\left\vert \chi_{E\cup E_{0}}-\chi_{E_{0}}\right\vert \,d\mu<\delta_1,\quad\int_{\Omega}\left\vert \chi_{E_{0}\setminus E}-\chi_{E_{0} }\right\vert \,d\mu<\delta_{1}, $$ and so, \begin{align*} \sup_{n,\,l\geq k}\left\vert \int_{E\cup E_{0}}f_{n}\,d\mu-\int_{E\cup E_{0} }f_{l}\,d\mu\right\vert & \leq\varepsilon,\\ \sup_{n,\,l\geq k}\left\vert \int_{E_{0}\setminus E}f_{n}\,d\mu-\int _{E_{0}\setminus E}f_{l}\,d\mu\right\vert & \leq\varepsilon. \end{align*} It follows that for any $n\geq k$, \begin{align*} \left\vert \int_{E}f_{n}\,d\mu\right\vert \leq & \left\vert \int_{E} f_{k}\,d\mu\right\vert +\left\vert \int_{E}f_{n}\,d\mu-\int_{E}f_{k} \,d\mu\right\vert \\ \leq & \int_{E}|f_{k}|\,d\mu+\left\vert \int_{E\cup E_{0}}f_{n}\,d\mu -\int_{E\cup E_{0}}f_{k}\,d\mu\right\vert \\ & +\left\vert \int_{E_{0}\setminus E}f_{n}\,d\mu-\int_{E_{0}\setminus E} f_{k}\,d\mu\right\vert \\ \leq & 3\varepsilon. \end{align*} In particular, by replacing $E$ first with $E\cap\{f_{n}\geq0\}$ and then with $E\cap\{f_{n}<0\}$ it follows that $\int_{E}(f_{n})^{+}\,d\mu\leq3\varepsilon$ and that $\int_{E}(f_{n})^{-}\,d\mu\leq3\varepsilon$. Hence, $\int_{E} |f_{n}|\,d\mu\leq3\varepsilon$.

  • $\begingroup$ Thank you @Gio67, I get it! $\endgroup$ – alexp9 Nov 27 '17 at 10:38

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