An increasing sequence of non-negative functions in $\mathcal{L}_{1}(X, \mu, \mathbb{R})$ is $\mathcal L_1$-Cauchy

Question

To avoid any ambiguity, I first present the related definitions:

Let $(X, \mathcal{A}, \mu)$ be a complete, $\sigma$-finite measure space and $(E,|\cdot|)$ a Banach space.

• We say $f \in E^{X}$ is $\boldsymbol{\mu}\textbf{-simple}$ if $f(X)$ is finite, $f^{-1}(e) \in \mathcal{A}$ for every $e \in E,$ and $\mu\left(f^{-1}(E \backslash\{0\})\right)<\infty$. We denote by $\boldsymbol{\mathcal{S}(X, \mu, E)}$ the set of all $\mu$-simple functions.

• Suppose $f_n, f \in E^{X}$ for $n \in \mathbb{N} .$ Then $(f_n)_{n \in \mathbb N}$ converges to $f$ $\boldsymbol{\mu}\textbf{-almost everywhere}$ if and only if there is a $\mu$-null set $N$ such that $f_{n}(x) \rightarrow f(x)$ for all $x \in N^{c}$.

• A function $f \in E^{X}$ is said to be $\boldsymbol{\mu}\textbf{-measurable}$ if there is a sequence $\left(f_{j}\right)$ in $\mathcal{S}(X, \mu, E)$ such that $f_{j} \rightarrow f$ $\mu$-almost everywhere as $j \rightarrow \infty$. We denote by $\boldsymbol{\mathcal{L}_0(X, \mu, E)}$ the set of all $\mu$-measurable functions.

• Let $\|\varphi\|_{1}:=\int_{X}|\varphi| \, d \mu$ for all $\varphi \in \mathcal{S}(X, \mu, E)$. Then $\|\cdot\|_{1}$ is a seminorm on $\mathcal{S}(X, \mu, E)$.

• In the following, we always provide the space $\mathcal{S}(X, \mu, E)$ with the topology induced by $\|\cdot\|_{1}$. Then we may also call a Cauchy sequence in $\mathcal{S}(X, \mu, E)$ an $\boldsymbol{\mathcal{L}_{1}}\textbf{-Cauchy sequence}$.

• A function $f \in E^{X}$ is called $\boldsymbol{\mu}\textbf{-integrable}$ if $f$ is a $\mu$-a.e. limit of some $\mathcal{L}_{1}$-Cauchy sequence $\left(\varphi_{j}\right)$ in $\mathcal{S}(X, \mu, E)$. We denote the set of $E$-valued, $\mu$-integrable functions of $X$ by $\boldsymbol{\mathcal{L}_{1}(X, \mu, E)}$.

• After these preparations, we define the integral of integrable functions in a natural way, extending the integral of simple functions. Suppose $f \in \mathcal{L}_{1}(X, \mu, E)$. Then there is an $\mathcal{L}_{1}$-Cauchy sequence $\left(\varphi_{j}\right)$ in $\mathcal{S}(X, \mu, E)$ such that $\varphi_{j} \rightarrow f \quad \mu$-a.e. The quantity $$\int_{X} f \, d \mu:=\lim _{j} \int_{X} \varphi_{j} \, d \mu$$ exists in $E$, and is independent of the sequence $\left(\varphi_{j}\right)$. This is called the Bochner-Lebesgue integral of $f$ over $X$ with respect to the measure $\mu$.

• For $f \in \mathcal{L}_{1}(X, \mu, E)$, let $\|f\|_{1}:=\int_{X}|f| \, d \mu$. Then $\|\cdot\|_{1}$ is a seminorm on $\mathcal{L}_{1}(X, \mu, E),$ called the $\mathcal{L}_{1}$-seminorm. We will always give $\mathcal{L}_{1}(X, \mu, E)$ the topology induced by the seminorm $\|\cdot\|_{1}$.

Then I have the following problem:

Suppose $\left(f_{n}\right)$ is an increasing sequence in $\mathcal{L}_{1}(X, \mu, \mathbb{R})$ such that $f_{n} \geq 0,$ and suppose it converges $\mu$-a.e. to $f \in \mathcal{L}_{1}(X, \mu, \mathbb{R})$. Show that $\left(f_{n}\right)$ is a $\mathcal L_1$-Cauchy sequence in $\mathcal{L}_{1}(X, \mu, \mathbb{R})$

My attempt:

Assume that there is a $\mu$-null set $N$ such that $f_{n}(x) \rightarrow f(x)$ for all $x \in N^{c}$.

Fix an $\epsilon >0$. Then for each $x \in N^c$, there is $M_x \in \mathbb N$ such that $f(x) - f_n(c) < \epsilon$ for all $n > M_x$.

Then I'm stuck because $M_x$ depends on $x$.

Could you please leave me some hints to finish the proof? Thank you so much!

Answer 1

Since $(f_n)$ is an increasing sequence, the sequence $$a_n = \int_X f_n\,d\mu$$ is also increasing. And for $m > n$ we have $$\lVert f_m - f_n\rVert_1 = \int_X \lvert f_m - f_n\rvert\,d\mu = \int_X f_m - f_n\,d\mu = a_m - a_n\,.$$ Thus $(f_n)$ is an $\mathcal{L}_1$-Cauchy sequence if and only if $(a_n)$ is a Cauchy sequence in $\mathbb{R}$. A monotonic sequence in $\mathbb{R}$ is a Cauchy sequence if and only if it is bounded.

Since $f_n \leqslant f$ for all $n$ and $f \in \mathcal{L}_1$ we have $$a_n = \int_X f_n\,d\mu \leqslant \int_X f\,d\mu < +\infty$$ and thus $(a_n)$ is bounded (since the sequence is increasing it suffices to exhibit an upper bound, but the lower bound $a_n \geqslant 0$ is here trivial too), hence a Cauchy sequence. By the above, $(f_n)$ is a Cauchy sequence.

Conversely, if $(f_n)$ is a Cauchy sequence in $\mathcal{L}_1$, then we can find simple functions $\varphi_n$ such that $\lVert \varphi_n - f_n\rVert_1 < 2^{-n}$ for all $n$, whence $(\varphi_n)$ is an $\mathcal{L}_1$-Cauchy sequence in $\mathcal{S}(X,\mu,\mathbb{R})$, and $\varphi_n(x) \to f(x)$ whenever $f_n(x) \to f(x)$. Thus $f$ is an almost-everywhere limit of an $\mathcal{L}_1$-Cauchy sequence in $\mathcal{S}(X,\mu,\mathbb{R})$, consequently $f \in \mathcal{L}_1(X,\mu,\mathbb{R})$.

Answer 2

By Monotone Convergence Theorem $\int f_n \to \int f$. By DC T $\int (f-f_n)^{+} \to 0$ since $ 0 \leq (f-f_n)^{+} \leq f$ and $f$ is integrable. Now $\int (f-f_n)^{-} =-\int (f_n -f) +\int (f-f_n)^{+} \to 0$ s0 $\int |f_n-f| =\int (f-f_n)^{+} +\int (f-f_n)^{-} \to 0$. Finally $\int |f_n-f_m| \leq \int |f_n-f|+\int |f_m-f| \to 0$.

Answer 3

I've just figured out a proof, so I posted it here. It would be great if someone helps me verify it. Thank you so much!

---

Assume the contrary that $(f_n)$ is not a $\mathcal L_1$-Cauchy sequence in $\mathcal{L}_{1}(X, \mu, \mathbb{R})$. Then there is $\epsilon >0$ such that $\forall N >0, \exists M > N: \|f_{M+1} - f_M \|_1 \ge \epsilon$. Then we can extract a subsequence $(f_{\psi (n)})$ such that $\| f_{\psi (n+1)} - f_{\psi (n)}\|_1 \ge \epsilon$. This means $$\forall n \in \mathbb N: \int ( f_{\psi (n+1)} - f_{\psi (n)} ) \, d \mu \ge \epsilon$$

On the other hand, $$\begin{aligned} &\int f \, d \mu \\ \ge &\int f_{\psi (n+1)} \, d \mu \\ = &\int \left ( f_{\psi (0)}+ \sum_{k=0}^n ( f_{\psi (k+1)} - f_{\psi (k)} ) \right ) d \mu \\ = &\int \left ( f_{\psi (0)}+ \sum_{k=0}^n ( f_{\psi (k+1)} - f_{\psi (k)} ) \right ) d \mu \\ = &\int f_{\psi (0)} \, d \mu+ \sum_{k=0}^n \int ( f_{\psi (k+1)} - f_{\psi (k)} ) d \mu \\ \ge &\int f_{\psi (0)} \, d \mu + n \epsilon \end{aligned}$$

Taking the limit $n \to \infty$, we get $$\int f \, d \mu \ge \infty$$

This is a contradiction. As such, $(f_n)$ is not a $\mathcal L_1$-Cauchy sequence in $\mathcal{L}_{1}(X, \mu, \mathbb{R})$.