1
$\begingroup$

Background Information:

Proposition 2.16 - If $f\in L^+$, then $\int f = 0$ iff $f = 0$ a.e.

Proof - Suppose $f = \sum_{j}a_j\chi_{E_j}$, then $\int f = 0$ iff $a_j = 0$ or $\mu(E_j) = 0$. In general, if $f = 0$ a.e., and $\phi$ is a simple function with $0\leq\phi\leq f$ then $\phi = 0$ a.e. and we have $$\int f = \sup_{\phi\leq f}\int \phi = 0$$ Conversely, suppose $E_n = \{f\geq \frac{1}{n}\}$, then $$\int f \geq \int \frac{1}{n}\chi_{E_n} = \frac{1}{n}\mu(E_n)$$ So, $$\mu(E_n)\leq n\int f = 0$$ then $$\mu(\{f\geq 0\}) = \mu\left(\bigcup_{n}E_n\right) = 0$$

I am pretty sure this proof is correct. My question now pertains to the Corollary of this proposition:

Question:

Corollary 2.17 - If $\{f_n\}\subset L^+$, and $f_n(x)$ increases to $f(x)$ for a.e. $x$, then $\int f = \lim_{n\rightarrow \infty}\int f_n$.

My idea: Let $f = \sum_{1}^{n}a_j\chi_{E_j}$ then $\{E_j\}_{j}$ is a finite disjoint family of measurable functions such that $X = \bigcup_{j}E_j$. We have that $f_n(x)$ increases to $f(x)$ for a.e. $x$ this means that there is a measurable set $F$ such that $x\in F$ with $\mu(F) = 0$. Then $f_n$ increases to $f$ on $X\setminus E$ and then applying the Monotone Convergence theorem $$\int \chi_{X\setminus E}f_n\rightarrow \int \chi_{X\setminus E} f = \int f$$

My apologies if this is a bit messy. Any suggestions is greatly appreciated.

$\endgroup$
8
  • $\begingroup$ I assume the notation $L^+$ refers to the set of measurable functions from $X$ to $[0,\infty]$ (or maybe $[0,\infty)$), where $X$ is some measure space. Is this correct? If so, then why are you assuming that $f$ can be written in the form $\sum_{1}^{n}a_j \chi_{E_j}$? In other words, you seem to be assuming that $f$ is a simple function. $\endgroup$ – Bungo Jun 23 '16 at 22:23
  • $\begingroup$ Yes $L^+$ is the set of measurable functions from $X$ to $[0,\infty]$. I am letting $f$ be a simple function, I don't think there is anything wrong with that. But I could be wrong $\endgroup$ – Wolfy Jun 23 '16 at 22:25
  • $\begingroup$ But a sequence $(f_n)$ of arbitrary elements of $L^+$ will not generally converge to a simple function, not even if all of the $f_n$'s are simple functions. $\endgroup$ – Bungo Jun 23 '16 at 22:27
  • $\begingroup$ P.S. you are citing the monotone convergence theorem in your proof, but isn't that exactly what corollary 2.17 is? $\endgroup$ – Bungo Jun 23 '16 at 22:28
  • $\begingroup$ I see your point, Folland has MCT as Theorem 14, I made a post on that earlier. $\endgroup$ – Wolfy Jun 23 '16 at 22:30
2
$\begingroup$

Note: your proof of 2.17 is incorrect, because you assume that an arbitrary element $f \in L^+$ can be written as $f = \sum_{j=1}^{\infty}a_j \chi_{E_j}$, but in fact this is only true of simple functions. Note that nonnegative simple functions are elements of $L^+$, but a general element of $L^+$ is any measurable function from $X$ to $[0,\infty]$.

We wish to prove Corollary 2.17, which is as follows.

Corollary 2.17: If $f_n$ is a sequence of functions in $L^+$, and $f \in L^+$, and $f_n(x)$ increases to $f(x)$ for almost every $x$, then $\int f = \lim \int f_n$.

This is very similar to Theorem 2.14, the Monotone Convergence Theorem, which states:

Theorem 2.14: If $f_n$ is a sequence of functions in $L^+$, and $f_j \leq f_{j+1}$ for all $j$, and $f = \lim f_n (= \sup f_n)$, then $\int f = \lim \int f_n$.

The only difference is that 2.14 assumes convergence everywhere, whereas 2.17 relaxes this to convergence almost everywhere. Let's see how to use Theorem 2.17 to prove Corollary 2.17.

We will also use Proposition 2.16, which states:

Proposition 2.16: If $f \in L^+$, then $\int f = 0$ iff $f = 0$ almost everywhere.

Proof of Corollary 2.17

We are given that $f_n \in L^+$ and $f \in L^+$, and $f_n(x)$ increases to $f(x)$ for almost every $x$. So, let $E$ denote the set of those $x$ for which $f_n(x)$ does not increase to $f(x)$. Then $E$ has measure zero (by definition of "almost everywhere"), and $f_n(x)$ increases to $f(x)$ for every element of $X \setminus E$. Therefore, $$\int_{X \setminus E}f = \int_X f \chi_{X \setminus E} = \lim \int_X f_n \chi_{X \setminus E} = \lim \int_{X \setminus E}f_n\quad\quad(*)$$ where the middle equality holds by the Monotone Convergence Theorem (2.14) since $f_n \chi_{X \setminus E}(x)$ increases to $f \chi_{X \setminus E}(x)$ for every $x \in X$.

Now, note that $\int_E f = \int_X f \chi_E = 0$ by Proposition 2.16 (since $f \chi_E = 0$ almost everywhere). Similarly, $\int_E f_n = 0$. Equation $(*)$ remains true if we add zero to both sides, so: $$\begin{aligned} \int_X f &= \underbrace{\int_{E} f}_{=0} + \int_{X \setminus E}f \\ &= \lim\underbrace{\int_{E}f_n}_{=0} + \lim\int_{X \setminus E}f_n \\ &= \lim\left(\int_{E}f_n + \int_{X \setminus E}f_n \right) \\ &= \lim \int_X f_n \\ \end{aligned} $$

$\endgroup$
11
  • $\begingroup$ Actually, Folland takes the same approach in making $f$ simple. He could of made a mistake though... $\endgroup$ – Wolfy Jun 23 '16 at 23:29
  • $\begingroup$ @Wolfy In which proof? I don't see him making that assumption in the proof of 2.17. $\endgroup$ – Bungo Jun 23 '16 at 23:30
  • $\begingroup$ If you look at proposition 2.16, he has $f$ being a simple function $\endgroup$ – Wolfy Jun 23 '16 at 23:32
  • $\begingroup$ @Wolfy: Right, in the first sentence he shows that it is true if $f$ is simple. But starting with the second sentence he assumes $f$ is a general element of $L^+$ and uses the fact that the integral of $f$ is defined as the supremum of the integrals of simple functions $\phi$ such that $\phi \leq f$. (Now that I read your proof of 2.16 more carefully, I see that you are doing the same thing.) $\endgroup$ – Bungo Jun 23 '16 at 23:34
  • $\begingroup$ Correct, and I did pretty much the same general case so I am not sure why my proposition 2.16 $\endgroup$ – Wolfy Jun 23 '16 at 23:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.