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MCT: Let $\{f_n\}_{n=1}^{\infty}$ be a sequence of non-negative measurable functions that satisfy $f_n(x) \le f_{n+1}(x)$ a.e. $x$ and $\lim_{n \rightarrow \infty}f_n(x) = f(x)$ a.e. $x$. Then, $\lim_{x \rightarrow \infty} \int f_n = \int f$.

Assume that we are given the following function: (Note that $g_k$ is a measurable function.)

$g_n(x) = |g_1(x)| + \sum_{k=1}^{n} {|g_{k+1}(x) - g_{k}(x)|}$.

We define

$g(x) = |g_1(x)| + \sum_{k=1}^{\infty} {|g_{k+1}(x) - g_{k}(x)|}$.

Suppose that we do not know if $g_n(x)$ converges as $n \rightarrow \infty$. Then, can we still apply MCT to the function $g_n$?

That is, I am not sure if the condition $\lim_{n \rightarrow \infty}g_n(x) = g(x)$ a.e. $x$ can be considered to be satisfied if the series defining $g$ is not known to be bounded. I am thinking that usually when we write $\lim_{n \rightarrow \infty}g_n(x) = g(x)$, we are saying for that $\forall \epsilon > 0, \exists N$ s.t. $|g_n(x)-g(x)| < \epsilon$ if $n > N$. This won't make sense if the series diverges.

I am asking this question because Stein used MCT to prove the Riesz-Fischer Theorem in his third book (pg 70) and in that case, he applied MCT on a series that might not converge.

In addition to this, is it possible to apply MCT to $g^2$?

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First of all, note that each function $f_n$ is measurable and non-negative. Moreover, the sequence $(f_n)$ is non-decreasing. Let us now fix $x$. Then either $(f_n(x))_n$ is bounded and so the sequence $(f_n(x))_n$ does converge to some finite number $f(x)$ or $(f_n(x))_n$ is unbounded. In the latter case, $(f_n(x))$ tends to $+ \infty$ and we can define $f(x) = + \infty$. In both cases, $f$ is a non-negative and measurable function taking values in $[0, \infty]$. Thus, we can apply the MCT.

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  • $\begingroup$ I think what you said is true. My main question is actually the following. What does $\lim_{n \rightarrow \infty}g_n(x) = g(x)$ actually mean when the limit is not well-defined? Suppose the limit is not well-defined, is it true that $\lim_{n \rightarrow \infty}g_n^2(x) = g^2(x)$? $\endgroup$ – Student Apr 16 '16 at 0:34
  • $\begingroup$ What do you mean by "when the limit is not well-defined"? The limit $f$ is well-defined in $[0, \infty]$. $\endgroup$ – Mike Apr 16 '16 at 7:10
  • $\begingroup$ For example, when $g_n(x) = (-1)^n$. Of course in my example above the limit is well-defined because $g_n$ is a non-negative increasing function. But suppose that in some other case we have $g_n(x) = (-1)^n$, does it make sense to write $\lim_{n \rightarrow \infty}{g_n(x)} = g(x)$? $\endgroup$ – Student Apr 16 '16 at 17:42
  • $\begingroup$ No, it doesn't make any sense to write $\lim_{n\to \infty}g_n (x)$ if $g_n(x)$ is given by $g_n(x) = (-1)^n$ since $\limsup_{n \to \infty}g_n(x) = 1 \neq -1 = \liminf_{n \to \infty} g_n(x)$. $\endgroup$ – Mike Apr 16 '16 at 18:42

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