Monotone increasing sequence in Lp convergent a.e.

Question

I´m having trouble with the proof of the following theorem in measure theory:

Consider the sequence $(f_n)_{\,n \in \Bbb N}\subset L^p(X)$ with $\;0\le f_n(x)\le f_{n+1}(x) \;\;\;\forall x\in X\;\forall n\in \Bbb N$

and the property that there exists a constant $C \in \Bbb R$ such that $\|f_n\|_p \le C \;\;\; \forall n \in \Bbb N$.

Than $f_n$ converges pointwise almost everywhere.

Or equivalently $E:=\{x\in X\;|\;f_n(x)\; does\; not\; converge \}$ is a null-set. Here is my attempt:

Suppose $E$ is not a null-set, than $E$ is either measurable but $\mu (E)\neq 0$ or $E$ is not measurable. Now I could show that the former leads to a contradiction but I dont know what to do for the case that $E$ is not measurable. Any Tips on how to go about this?

Answer 1

1. Monotonicity implies that $\lim_{n\to\infty} f_n(x)$ exists for every $x$. Denote this limit by $f(x)$. Note that $f(x)$ may be $+\infty$ for some $x$.

2. Let $E$ denote the set $\{x:f(x)=\infty\}$. Is this measurable ? Yes.

Observe that $\cup_{n=1}^\infty \{x:f_n (x)> M\}\subset \{x:f (x)> M\}$. On the other hand if $f(x)>M$, then from $\lim_{n\to\infty} f_n (x) =f (x)$, it follows that $ f_n (x) > M$ for some large enough $n$, otherwise if $ f_n (x) \leq M \ \forall \ n$ then $\Rightarrow f(x) \leq M$ which is a contradiction. We proved:

$$\{x:f(x)>M\}= \cup_{n=1}^{\infty} \{x:f_n (x)>M\} \quad\quad (*)$$

The righthand side is measurable. Therefore so is the lefthand side.

Now

$$E=\{x:f(x) = +\infty\}= \cap_{M=1}^{\infty} \{x:f(x) >M\},$$

therefore $E$ is measurable.

3. Lastly, we show $\mu(E)=0$.

Note: all the above only relied on the assumption that $(f_n)$ is a nondecreasing sequence of nonnegative measurable functions.

Here we need to employ the additional assumption $\sup_{n\in N} \|f_n\|_p <C$.

Clearly, for every $M$, $E\subset \{x:f(x) > M\}$, and by $(*)$, the latter set is $\cup_{n=1}^{\infty} A_n$ where $A_n =\{f_n (x) > M\}$. However, since $f_n \le f_{n+1}$, $A_n \subset A_{n+1}$. Therefore by continuity of measure with respect to nondecreasing sequences of sets,

$$ \mu (\cup_{n=1}^{\infty} A_n) =\lim_{n\to\infty} \mu (A_n).$$

Now apply Chebychev's inequality:

$$\mu(A_n) \le \frac{ \int_{A_n} f_n^p d\mu}{M^p}\le \int f_n ^p d \mu / M^p\le C^p/M^p.$$

(the first inequality follows because because on $A_n$, $f_n^p>M^p$).

Putting it all together, we have

$$ \mu (E) \le \mu ( \{x:f(x) > M\})=\lim_{n\to\infty} \mu (A_n) \le \frac{C^p}{M^p}.$$

This is true for every $M$. Therefore letting $M\to\infty$, the result follows.

Answer 2

The sequence $f_n$ is increasing, so $f_n$ will, at each point, either converge (to a finite number) or go off to infinity. The set $E$ you have defined is measurable because $E=\{ x\in X\;| \limsup\limits_{n\to\infty} f_n (x) = \infty\}$ and $\limsup\limits_{n\to\infty} f_n$ is well known to be a measurable function when the $f_n$ are measurable.