Use Fatou Lemma to show that $f$ takes real values almost everywhere.

Question

Let $(f_n)$ be a sequence in $L^p$ such that for each positive integer $n$, $ \| f_{n+1}-f_n\|_{p} <\frac 1{2^n} $. Define $f: X \to [0,\infty]$ with $$ f(x)= \sum_{n=1}^\infty| f_{n+1}(x)-f_n(x)|. $$ Use Fatou Lemma to show that $f$ takes real values almost everywhere.

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My thoughts: We must show the existence of $A$ such that $\mu(A)=0$ and for all $x\in A^C$, $f\in\mathbb R$ using Fatous Lemma which states $$ \int_X\liminf_nf_n \le\liminf_n\int_X f_n. $$

My idea taking pieces of the completeness proof of $L^p$ is that we have that the series $f(x)=\displaystyle\sum_{n=1}^\infty\vert f_{n+1}(x)-f_n(x)\vert$ converges so $f(x)=\displaystyle\sum_{n=1}^\infty\vert f_{n+1}(x)-f_n(x)\vert<\infty$ and this holds for all $x\in X$ and then how will I define the set $A$?

I am not using at all Fatou Lemma because I don't know how.

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[Added later:]

I would also like to understand steps of the proof in Alex R.'s answer below.

It is shown that

$$\int_A|f(x)|^pd\mu\leq \liminf_m\int_A|F_m(x)|^p dx\leq\liminf_m\sum_{n=1}^m\frac{1}{2^{pn}}d\mu<\infty.$$

There are a lot of missing steps. I tried to fill.

$\displaystyle\int_A\vert f(x)\vert^pdx =\int_A|\liminf_mF_m(x)|^pdx=\int_A\liminf_m|F_m(x)|^pdx\le\liminf_m\int_A|F_m(x)|^pdx =\liminf_m\int|\sum^m_{n=1}|f_{n+1}(x)-f_n(x)||^pdx \fbox{=}\liminf_m\sum^m_{n=1}\int_A|f_{n+1}(x)-f_n(x)|^pdx =\liminf_m\sum^m_{n=1}||f_{n+1}(x)-f_n(x)||_p^p\le\liminf_m\sum^m_{n=1}\frac{1}{2^{np}}\fbox=?$

Question1: Assuming the equalities and inequalities are correct, Why is the second equality, the boxed equality and what is equal in the last boxed equality?

It is also shown that

Taking $A$ to be $\mathbb{R}$, it follows that $\int_A|f(x)|^pd\mu<\infty$ which implies $\mu(\{|f|=\infty\})=0$.

Question2: Should not have been written as

Take $A$ to be $\mathbb R.$

From all the arguments above (and not only $A=\mathbb R$), it follows that $\int_A|f(x)|^pd\mu<\infty$ which implies $\mu(\{|f|=\infty\})=0$ ??

Seems like as it originally stands, the affirmation depends only on $A=\mathbb R$. But it is independent of taking $A$ as $\mathbb R$. Am I correct?

Answer 1

I would like to directly answer your original question (for $1\le p< \infty$):

Let $(f_n)$ be a sequence in $L^p$ such that $$ \| f_{n+1}-f_n\|_{p} <\frac 1{2^n} \quad \forall n\in\mathbb{N}. $$ and define $f: X \to [0,\infty]$ with $$ f(x):= \sum_{n=1}^\infty| f_{n+1}(x)-f_n(x)|.\tag{1} $$ Use Fatou Lemma to show that $f$ takes real values almost everywhere.

For each positive integer $N$, define $F_N:X\to[0,\infty]$ with $F_N(x)=\sum_{n=1}^N|f_{n+1}(x)-f_n(x)|.$ Then for each $x\in X$, $$ f(x)=\lim_{N\to\infty}F_N(x) $$ by the definition of $f$, and thus by continuity of the function $z\mapsto z^p$, for each $x\in X$, $$ |f(x)|^p=\big(\lim_{N\to\infty}F_N(x)\big)^p=\lim_N |F_N(x)|^p. $$

Now by Fatou lemma, $$ \int |f|^p\,d\mu\le \liminf_N \int |F_N|^p.\tag{2} $$ But by the Minkowski inequality, for each $N$, $$ \int |F_N|^p= \|F_N\|_p^p\le (\sum_{n=1}^N\|f_{n+1}-f_n\|_p)^p\le(\sum_{n=1}^N\frac{1}{2^{n}})^p \tag{3} $$

Taking $\liminf_{N\to\infty}$ in (3) and then applying (2), one gets $$ \int |f|^p<\infty $$ and thus $\mu\{x\in X:|f(x)|^p=\infty\}=0$ and hence¹ $\mu\{x\in X:|f(x)|=\infty\}=0$.

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¹ See, e.g. this standard result: An integrable Functions is almost everywhere finite

Answer 2

Let $F_m(x)=\sum_{i=1}^m|f_{n+1}(x)-f_n(x)|$. Since $F_m(x)$ is increasing for each $x$, its limit in $m$ necessarily exists in the extended reals. So $\lim_{m\rightarrow\infty}F_m(x)=\liminf_{m\rightarrow\infty}F_m(x)=f(x)$

Then by Fatou:

$$\int_A|f(x)|^pd\mu\leq \liminf_m\int_A|F_m(x)|^p dx\leq\liminf_m\sum_{n=1}^m\frac{1}{2^{pn}}d\mu<\infty.$$

Taking $A$ to be $\mathbb{R}$, it follows that $\int_A|f(x)|^pd\mu<\infty$ which implies $\mu(\{|f|=\infty\})=0$.

Answer 3

Hint: Since $\Vert f_{n+1}-f_n\Vert_p<\frac 1{2^n}$, if $g_k(x)=\displaystyle\sum_{n=1}^k\vert f_{n+1}(x)-f_n(x)\vert$ , then Minkowski's inequality implies that $\|g_k\|_p\le 1.$ And now an application of Fatou to $(g^p_k)$ shows that $\|f\|_p\le 1$, for the $f$ as in your question, and this implies that $f(x)<\infty$ almost everywhere.