Consider the measure space $(A,\mathcal{F},\mu)$. We say that a real measurable function $f$ on $A$ is integrable if $\int_A \mid f\mid d\mu < \infty$.
Furthermore, an integral of a real measurable function is $\int fd\mu=\int f^+-\int f^-d\mu$, and if $f\geq 0$ the integral is given by $\sup\{\int \phi d\mu:0\leq \phi\leq f, \phi \ \text{is simple function}\}$.
I know that if $f$ is integrable then $\int f d\mu<\infty$.
Questions
1) If $\int f d\mu<\infty$, then $f$ is integrable ?
2) If $f$ is integrable, then $\int_E f d\mu<\infty, \forall E\in\mathcal{F}$?
My attempts
2) $\infty>\int \mid f\mid d\mu= \int_{E}\mid f\mid d\mu+\int_{A\setminus E} \mid f \mid d\mu\geq \int_{E}\mid f\mid d\mu \geq \int_{E} f d\mu, \forall E\in \mathcal{F}.$
1) $\infty>\int f d\mu=\int f^+ d\mu-\int f^- d\mu$. I see that $\int \mid f\mid d\mu$ is an upper bound for the integral of $f$, but I can't see why it must be finite. Can someone help me?
UPDATE: if $ \int f^- d\mu$ is infinite and $\int f^+ d\mu$ finite , $\int f d\mu<\infty$ but $\int \mid f \mid d\mu$ is infinite! So the statement is false. In prticular, take $f(x)=-1-x^2$ on the Lebesgue measure space. What do you think?
Thnks!
Best Answer
Consider $f(x) = -1$ for all $x \in \mathbb{R}$.
People A say: $\int f dx = -\infty$.
People B say: $\int f dx$ does not exist!
Everyone agrees that: $-\infty < \infty$.
So it is reasonable for people A to say: $$ \int f dx = -\infty < \infty$$ On the other hand, it is reasonable for people B to say: $\int f dx$ is not less than $\infty$ because $\int f dx$ does not exist, and something that does not exist cannot be compared in an inequality statement!
Convention for People A
Let $(A, \mathcal{F}, \mu)$ be a measure space triplet. If $f:A\rightarrow\mathbb{R}$ then define \begin{align} f^+(x) &= \max[f(x),0] \quad \forall x \in A \\ f^-(x) &= -\min[f(x),0] \quad \forall x \in A \end{align} Then $f^+$ and $f^-$ are nonnegative functions and \begin{align} f(x) &= f^+(x) - f^-(x) \quad \forall x \in A\\ |f(x)| &= f^+(x) + f^-(x) \quad \forall x \in A \end{align} It can be shown that if $f$ is measurable then $f^+, f^-, |f|$ are nonnegative and measurable.
If $f$ is a nonnegative and measurable function then $\int fd\mu$ is always defined (possibly equal to $\infty$). Also it satisfies: $$ \int f d\mu = \lim_{M\rightarrow\infty} \int_0^M \mu(\{x \in A : f(x)\geq t\})dt $$ If $f$ is a measurable function (possibly taking negative values) then we define $$ \int f d\mu = \int f^+ d\mu - \int f^- d\mu$$ whenever the right-hand-side avoids the undefined case of $\infty - \infty$. That is, $\int f d\mu$ is defined if and only if either $\int f^+d\mu <\infty$ or $\int f^-d\mu < \infty$.
With this definition we observe that $$ \int |f|d\mu = \int f^+d\mu + \int f^-d\mu$$ and $\int |f|d\mu <\infty$ if and only if $\int f^+d\mu<\infty$ and $\int f^-d\mu<\infty$.
Convention for People B
Everything is the same, except that people B require all integrals $\int f d\mu$ to have finite values. So they say that $\int f d\mu$ is defined if and only if $\int f^+d\mu<\infty$ and $\int f^- d\mu<\infty$. That is, they say that $\int f d\mu$ is defined if and only if $\int |f|d\mu<\infty$.
The advantage of this convention is that it removes the cases when $\int f^+d\mu$ or $\int f^-d\mu$ are infinite, so there is often less work to do (we do not have to consider so many cases when proving things). However, people B routinely use the people A language. For example, people B will routinely use the Borel-Cantelli lemma by saying things like
but of course, strictly speaking, the equation $\sum_{n=1}^{\infty} P[A_i]=\infty$ is not allowed to exist as an equation under the people B convention: If the equation is true then the left-hand-side does not exist and so the equation is meaningless!
Now this Borel-Cantelli example uses a sum instead of an integral, but the people B convention for sums is (usually) the same as for integrals, and in fact a sum can be viewed as an integral under a certain measure.
One can also observe that the people B convention is problematic because it implies that the inequality $\int f d\mu < \infty$ is either meaningless or trivially true: If the integral $\int f d\mu$ is allowed to exist then (by the people B convention) it must be finite and so there is no reason to write such an inequality (it is trivially true). On the other hand if $\int f d\mu$ does not exist then we are not allowed to pretend that it exists in the inequality $\int fd\mu < \infty$. People B must live with these (minor) contradictions.
If you want the best of both worlds, being free of contradictions and also doing less work, you can just write the preamble "Suppose $\int |f|d\mu <\infty$" before your analysis of integrals of $f$. Under the condition $\int |f|d\mu <\infty$, people A and B can co-exist happily.