2
$\begingroup$

Let $f\in L^+(X,\mathscr{M},\mu)$ and $\int f <\infty$ for every $\epsilon>0$ there exists $E$ such that $\mu(E)<\infty$ and $\int_E f > \int f -\epsilon$.

Let $f\in L^+$, then we know there exists an increasing sequence of simple function $\{\phi_n\}$ where $\phi_n=\sum^{k_n}_{i=1} a_i \chi_{E_i}$ where $\bigsqcup^{k_n} E_i=X$, Let $\epsilon>0$ by MCT we know there exists $N>0$ such that $$\big|\int\phi_n-\int f\big|<\epsilon$$ for all $n>N$. Consider $N+1$, then we have $\int\phi_{N+1}>\int f -\epsilon$. Since $\phi_n \nearrow f$ then $\int f \geq \int \phi_n$.

I am not sure how to approch now to define a finite measurable set $E$, can someone give me some hint? I was thinking about using the disjoint set $E_i$ that made up the simple function, but then they union to the whole space by definition.

$\endgroup$
0

1 Answer 1

3
$\begingroup$

Rather than using the simple function approach, consider the level sets $E_M = \{x: f(x)\le M\}$ and the monotone convergence theorem applied to a particular sequence built out of $f$.


Added: The sets $E_M$ can have infinite measure owing to the small values of $f$. To complete this hint into an answer, we need to modify the sets $E_M$ to address the small values of $f$.

Spoiler for after you have a go at the adjustment:

Consider $E_{N,M} = \{x:N < f(x) \le M\} = E_N^\complement\cap E_M$. Given $\epsilon>0$, if $0 < N \le N(\epsilon)$ is sufficiently small, and $M(\epsilon)<M$ is sufficiently large, $\int_{E_{N,M}} f>\int f - \epsilon$. Moreover, by Tchebychev's inequality, $\mu(E_{N,M})\le \mu(E_N^\complement) \le \frac{1}{N}\int f < \infty$. Note how because $N(\epsilon)$ may be very small, the upper bound coming from Tchebychev may be very large, but still finite because $f$ is integrable. This is the only place we use the assumption $f$ is integrable.

$\endgroup$
4
  • $\begingroup$ Thanks! Would the sequence $f_n=f\chi_{E_n}$ work and then applied the same logic I did originally? Then $\int_{E_{N+1}}f=\int f_{N+1}>\int f -\epsilon$. Where $E_{N+1}=\{x:f(x)\leq N+1\}$ But then how would I argue that $E_{N+1}$ has finite measure? $\endgroup$
    – Remu X
    Jun 12, 2023 at 15:05
  • $\begingroup$ I was thinking using the fact that $m((0,M))=M<\infty $ then use this to argue its preimage under $f$ which is $E_M$ must be finite? $\endgroup$
    – Remu X
    Jun 12, 2023 at 15:41
  • $\begingroup$ @Remu: The sets $E_M$ can have infinite measure (think of a function like a Gaussian). You have to make some adjustment to address the small values of $f$. But since $f$ is integrable, you should be able to say something. $\endgroup$
    – Alex Ortiz
    Jun 12, 2023 at 15:49
  • 1
    $\begingroup$ @Remu: I added a clickable spoiler you can compare your answer with based on our discussion in the comments. $\endgroup$
    – Alex Ortiz
    Jun 12, 2023 at 16:26

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .