5
$\begingroup$

Let $(X, \mathcal A, \mu)$ be a measure space. Let $f:X \rightarrow [0, \infty]$ be a non-negative extended real-valued function. It is sometimes useful to consider the so-called upper integral $\int^* f d\mu$(for example we would like to define the outer measure of a subset $A$ of $X$ which should be the upper integral of the characteristic function of $A$). We define the upper integral $\int^* f d\mu$ as follows.

$\int^* f d\mu = \text{inf }\{\int g d\mu: f \le g, g:\text{ measurable}\}$.

It seems to me that the following proposition is correct.

Let $f_1\le f_2 \le \cdots$ be a non-decreasing sequence of functions $X\rightarrow [0, \infty]$. Let $f(x) = \text{lim}_{n\rightarrow\infty} f_n(x)$ for all $x\in X$. Then $\int^* f d\mu = \text{lim}_{n\rightarrow \infty} \int^* f_n d\mu$.

How do you prove this if it is correct?

Actually I think I have a proof of it, but I am not 100% sure that it is correct. I think it would be nice that someone confirms that I am on the right track.

$\endgroup$
2
  • $\begingroup$ Perhaps you could write your proof? $\endgroup$ Jul 14, 2015 at 5:30
  • $\begingroup$ @PrahladVaidyanathan Where can I write it? $\endgroup$
    – user254385
    Jul 15, 2015 at 0:09

3 Answers 3

0
$\begingroup$

If you assume that $\mu$ is $\sigma$-finite and define $f^*$ as a minimal measurable majorant of $f$ (which exists for $f:X \rightarrow \bar{\mathbb{R}}$ in these settings), then $f^*_n\uparrow f^*$ a.s. Consequently, using the ordinary monotone convergence theorem

$$\mathbb{E}^*f_n=\mathbb{E}f_n^*\uparrow \mathbb{E}f^*=\mathbb{E}^*f$$


  • $f^*$ is defined as: (1) $f^*\ge f$ and (2) for any measurable $g:X\rightarrow \bar{\mathbb{R}}$ with $g\ge f$, $f^*\le g$ a.s.
  • Convergence of $f_n^*$ follows from

$$f=\liminf f_n\le \liminf f_n^* \le \limsup f_n^* \le f^*$$

and the fact that $f^*$ is minimal measurable envelope of $f$.

$\endgroup$
3
  • $\begingroup$ Could you explain how you prove such $f^*$ exists? By the way, what is $\mathbb P^*A$? $\endgroup$
    – user254385
    Jul 15, 2015 at 0:03
  • $\begingroup$ Please read math.stackexchange.com/questions/1361495/… $\endgroup$
    – user254385
    Jul 15, 2015 at 20:00
  • $\begingroup$ @billford: Good point. I implicitly assumed a probability space... $\endgroup$
    – user140541
    Jul 15, 2015 at 20:06
0
$\begingroup$

Since nobody has posted a correct answer yet, I have decided to post my proof of which I am not 100% sure. I welcome any corrections.

Since $f_n \le f$ for all $n$, $\text{lim}_n \int^* f_n d\mu \le \int^* f d\mu$. Hence it suffices to prove that $\int^* f d\mu \le \text{lim}_n \int^* f_n d\mu$. If $\int^* f_n d\mu = \infty$ for some $n$, then the inequality clearly holds. Hence we may assume $\int^* f_n d\mu \lt \infty$ for all $n$.

Choose $\epsilon \gt 0$. We will show that there exists an increasing sequence of measurable functions $g_1\le g_2\le \cdots$ such that $f_n \le g_n$ and $\int g_n d\mu \le \int^* f_n d\mu + \epsilon$ for all $n$. Then by Lebesgue Monotone Convergence Theorem, $\int^* f d\mu \le \int g d\mu = \text{lim}_n \int g_n d\mu \le \text{lim}_n \int^* f_n d\mu + \epsilon$, where $g(x) = \text{lim}_n g_n(x)$ for all $x\in X$. Letting $\epsilon \rightarrow 0$, we have $\int^* f d\mu \le \text{lim}_n \int^* f_n d\mu$ as desired.

By the definition of the upper integral, there exists a measurable function $h_n$ such that $f_n\le h_n$ and $\int h_n d\mu \lt \int^* f_n d\mu + \epsilon/2^n$ for each $n$. Let $g_n = \text{sup}(h_1,\cdots, h_n)$. Since $f_n \le g_n,g_1\le g_2\le\cdots$, it suffices to prove

$\int g_n d\mu \le \int^* f_n d\mu + \epsilon(1 - 1/2^n)$ for all $n$.

We prove this by induction on $n$. The assertion is clear when $n = 1$. Since $g_{n+1} +\text{inf}(g_n, h_{n+1}) = \text{sup}(g_n, h_{n+1}) + \text{inf}(g_n, h_{n+1}) = g_n + h_{n+1}$ and $f_n \le \text{inf}(g_n, h_{n+1})$, we have

$\int g_{n+1} d\mu = \int g_n d\mu + \int h_{n+1} d\mu - \int \text{inf}(g_n, h_{n+1}) d\mu \le \int g_n d\mu + \int h_{n+1} d\mu - \int^* f_n d\mu\le \int^* f_{n+1} d\mu + \epsilon(1 - 1/2^n) + \epsilon/2^{n+1} = \int^* f_{n+1} d\mu +\epsilon(1 - 1/2^{n+1}).$

This completes the proof.

$\endgroup$
0
$\begingroup$

Lemma:(monotone convergence for Lebesgue of upper integral) Let $(X; \mu)$ be a measure space, and $f_n: X \to [0,\infty]$ a monotone sequence of not necessarily measurable functions, i.e. $$ 0 \leq f_1(x) \leq f_2(x) \leq \cdots \quad \text{for \, $\mu$-a.e.} \quad x \in X \, . $$ Let $f(x):=\lim_{n \to \infty} f_n(x)$. Then $$ \lim_{n\to \infty} \int^* f_n \ d\mu = \int^* f \, d\mu \, . $$

Proof. First the case $\int^* f d\mu < \infty$. Since outside a negligible set the sequence $f_n(x)$ is increasing in $n$, the limit does exist in $[0,\infty]$. So, $f$ is defined $\mu$-a.e. and furthermore $$ \forall n\in\mathbb{N}\, : \int^* f_n \leq \int^* f_{n+1} \leq \int^* f < \infty \, , $$ by monotonicity of upper integral (even for non-measurable functions.) $$ \limsup_{n\to \infty} \int^* f_n = \lim_{n \to \infty} \int^* f_n \leq \int^* f \, . $$ Since each $\int^* f_n < \infty$, by definition, there exist simple functions $\phi_n$ such that $$ 0 \leq f_n (x) \leq \phi_n (x) , \quad \mu-\text{a.e.} \, $$ and $$ \int^* f_n \leq \int \phi_n < \int^* f_n + 2^{-n} $$ Since for any $n \leq m$ (positive integers) $$ f_n \leq f_{m} \leq \phi_{m} , \quad \mu-\text{a.e.} $$ the measurable functions $$ \psi_n : = \inf_{m\geq n} \phi_m \, , \quad \psi:=\lim_{n\to \infty} \psi_n $$ satisfy, $\mu$-a.e., and for all $n \in \mathbb{N}$, the following $$ 0 \leq \psi_1 \leq \psi_2 \leq \cdots \, , $$ $$ 0 \leq f_n \leq \psi_n \leq \phi_n \, , $$ and $$ 0 \leq f \leq \psi \, . $$ Now, the monotone convergence theorem applied to the measurable $\psi_n$, together with the inequalities above, yield $$ \int^* f \leq \int \psi = \lim_{n\to \infty} \int \psi_n \leq \lim_{n\to \infty} \int \phi_n = \lim_{n\to \infty} \int^* f_n \leq \int^* f\, . $$ $\Box$

$\endgroup$

You must log in to answer this question.