The measurable function sequence $f_n: X \to \mathbb{R}, n=1, 2, \dots$ is assumed to satisfy $0 \leq f_n \leq f_{n+1} \leq 1$ for each $n \in \mathbb{N}$.

The sequence $\{\mu_n\}_{n=1}^\infty$ is defined by the following and is assumed to be $\mu_n \in (0, \infty)$ for each $n \in \mathbb{N}$: $$ \mu_n = \int_X f_n\ d\mu. $$

The function sequence $g_n: X \to \mathbb{R}$ is defined below.

$$ g_1 = \frac{1}{\mu_1}f_1,\\ g_n = -\frac{1}{\mu_{n-1}}f_{n-1} + \frac{1}{\mu_n}f_n\ (\forall n \geq 2). $$

At this point, we would like to ask the following

  1. $\sum_{n=1}^\infty \int_X g_n\ d\mu = 1 \to$Solved!

  2. $\int_X \sum_{n=1}^\infty g_n\ d\mu$ if $\lim_{n\to\infty}f_n$ is integrable.

  3. $\int_X \sum_{n=1}^\infty g_n\ d\mu$ if $\lim_{n\to\infty}f_n$ is NOT integrable.

What I know

$$ \sum_{i=1}^n g_i = \frac{1}{\mu_1}f_1 - \frac{1}{\mu_1}f_1 + \frac{1}{\mu_2}f_2 - \dots - \frac{1}{\mu_{n-1}}f_{n-1} + \frac{1}{\mu_n}f_n = \frac{1}{\mu_n}f_n,\\ \sum_{n=1}^\infty g_n = \lim_{n\to\infty}\sum_{i=1}^n g_i = \lim_{n\to\infty} \frac{1}{\mu_n}f_n,\\ \forall n\in\mathbb{N}, f_n \leq f_{n+1} \therefore \int_X f_n d\mu \leq \int_X f_{n+1} d\mu \therefore \mu_n \leq \mu_{n+1}. $$ Therefore, $$ \int_X \sum_{n=1}^\infty g_n\ d\mu = \int_X \lim_{n\to\infty} \frac{1}{\mu_n}f_n = \cdots\ ? $$

We can't use the monotonic convergence theorem now, can we? When do we use the fact that $\lim_{n\to\infty}f_n$ is integrable?


2 Answers 2


[This answer is for an earlier verison of the question].

$\sum\limits_{n=1}^{N}g_n$ is a telscopic sum and its value is just $\frac {f_N} {\mu_N}$. Hence, $\sum\limits_{n=1}^{\infty}g_n=\frac f {\mu}$ where $f=\lim f_n$. By monotone convergenceTheorem we get $\int \sum\limits_{n=1}^{\infty}g_n d\mu=1$. By the same computation of partial sums we get $ \sum\limits_{n=1}^{\infty} \int g_n d\mu=1$.

Your mistake: $\int g_md \mu$ is actually $0$ for all $n>1$.

  • $\begingroup$ $f_n \leq f_{n+1} \Rightarrow \frac{1}{\mu_{n+1}} \leq \frac{1}{\mu_n}$. Why can we use the monotonic convergence theorem when it is not $\frac{1}{\mu_n}f_n \leq \frac{1}{\mu_{n+1}}f_{n+1}$? $\endgroup$
    – ytnb
    Commented Dec 13, 2022 at 10:53
  • $\begingroup$ You said $(\mu_n)$ is assumed to be $\mu$ for all $n$, right? @ytnb $\endgroup$ Commented Dec 13, 2022 at 12:31
  • $\begingroup$ $\mu$ is an error. It has been corrected, please check it. @geetha290krm $\endgroup$
    – ytnb
    Commented Dec 21, 2022 at 3:02

Notice that $$\sum^\infty_{k=1}\int_Xg_n\,d\mu=\lim_N\sum^N_{n=1}\int_Xg_n\,d\mu=\lim_N\int_X\Big(\sum^N_{n=1}g_n\Big)\,d\mu $$ Let $f_0=0$ and $\mu_0=1$ so that $g_n=\frac{f_n}{\mu_n}-\frac{f_{n-1}}{\mu_n}$ holds for all $n\in\mathbb{N}$.

  • Suppose $f=\lim_nf_n\in L_1(\mu)$, that is $\int f\,d\mu<\infty$ (monotone convergence implies that $\lim_n\mu_n=\int_Xf_n\,d\mu=\int_X\lim_nf_n\,d\mu=\int_Xf\,d\mu=:\mu_\infty$. The assumption $f\in L_1$ means that $\mu_\infty<\infty$).
    One the one hand, a simple telescopic sum argument yields \begin{align} \sum^N_{n=1}\int_Xg_n\,d\mu=\sum^N_{n=1}\int_X\Big(\frac{f_n}{\mu_n}-\frac{f_{n-1}}{\mu_{n-1}}\Big)\,d\mu=\int_X\frac{f_N}{\mu_N}\,d\mu-\int_X\frac{f_0}{\mu_0}\,d\mu=1 \end{align} One the other hand, another telescopic sum type of argument yields \begin{align} \sum^\infty_{n=1}g_n=\lim_N\sum^N_{n=1}g_n=\lim_N\Big(\frac{f_N}{\mu_N}-\frac{f_0}{\mu_0}\Big)=\lim_N\frac{f_N}{\mu_N}=\frac{f}{\mu_\infty} \end{align} where $\mu_\infty=\lim_N\int_Xf_n\,d\mu=\int_Xf\,d\mu$.
    Putting things together \begin{align} \sum^\infty_{n=1}\int_Xg_n\,d\mu&=\lim_N\int_X\sum^N_{n=1}g_n\,d\mu=\lim_N\sum^N_{n=1}\int_Xg_n\,d\mu =1\\ &=\int_X\frac{f}{\mu_\infty}\,d\mu=\int_X\lim_N\sum^N_{n=1}g_n\,d\mu=\int_X\big(\sum^\infty_{n=1}g_n\Big)\,d\mu \end{align}

  • If $\mu_\infty=\infty$, then $\sum_ng_n=0$, in which case we have that $$\int_X\sum_ng_n\,d\mu=0<\sum^\infty_{n=1}\int_Xg_n\,d\mu=1$$

  • $\begingroup$ $f = \lim_{n\to\infty}f_n \in L^1(\mu) := \{f: measurable; \int_X |f|d\mu < \infty\}$, isn't it? Is there no need to put an absolute value on $f$? $\endgroup$
    – ytnb
    Commented Dec 21, 2022 at 3:47
  • $\begingroup$ Or are you removing absolute values from the condition that $f_n$ is non-negative? $\endgroup$
    – ytnb
    Commented Dec 21, 2022 at 3:54
  • 1
    $\begingroup$ @ytnb: for your fist comment, there is no need of absolute values as your functions $f_n$ are nonnegative by assumption. As $0\leq f_n\leq f_{n+1}$, the limit can be moved inside the integral: $\lim_n\int_Xf_n=\int_X\lim_nf_n=\int_Xf$. $\endgroup$
    – Mittens
    Commented Dec 21, 2022 at 4:28
  • $\begingroup$ Does the fact that 1. and 2. coincide mean that we can use the termwise integral theorem? $\endgroup$
    – ytnb
    Commented Dec 21, 2022 at 12:47
  • $\begingroup$ @ytnb: yes, this happens only if $\mu_\infty=\int_Xf<\infty$, otherwise you have an inequality (second bullet in my posting). $\endgroup$
    – Mittens
    Commented Dec 21, 2022 at 15:15

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