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

Let $$f\in L^+(X,\mathscr{M},\mu)$$ 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.

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 $$M(\epsilon) is sufficiently large, $$\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.

• 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? Jun 12, 2023 at 15:05
• 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? Jun 12, 2023 at 15:41
• @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. Jun 12, 2023 at 15:49
• @Remu: I added a clickable spoiler you can compare your answer with based on our discussion in the comments. Jun 12, 2023 at 16:26