I have recently been introduced with integrals for measurable functions. I have the following definitions:

Definition 1: For a measurable functions $f:S\to[0,\infty]$ and a sequence of simple functions $(f_n)_{n\geq 1}$ with $0\leq f_n\uparrow f$ we define the integral of $f$ over $E\in\mathcal{A}$ as $$\int_E fd\mu = \lim\limits_{n\to\infty}\int_E f_n d\mu, \text{ in } [0,\infty],$$ where $\mathcal{A}$ is the $\sigma$-algebra of $S$.

Definition 2: A measurable function $f:S\to\overline{\mathbb{R}}$ is called integrable when both $\int_S f^{+}d\mu$ and $\int_S f^{-}d\mu$ are finite. In this case the integral of $f$ over $E\in\mathcal{A}$ is defined as $$\int_E fd\mu = \int_E f^{+}d\mu - \int_E f^{-}d\mu.$$

I have to solve the following exercise:

Exercise: Let $f:S\to[0,\infty]$ be a integrable function. Show that $f<\infty$ almost everywhere.

What I think I should do: I think that I need to show that $\int_S f^{+}d\mu$ and $\int_S f^{-}d\mu$ finite implies that $f<\infty$ almost everywhere.


  • How do I show that $\int_S f^{+}d\mu$ and $\int_S f^{-}d\mu$ finite implies that $f<\infty$ almost everywhere? I'm particularly confused by the part almost everywhere. If $f \nless \infty$ somewhere, wouldn't this imply that $\int_S f^{+}d\mu$ is infinite?

  • How do integrals for measurable functions differ from the Riemann integral? I've learned about the Riemann integral before and I don't really understand what the difference is.

Thanks in advance!

  • $\begingroup$ It's important to know $0\cdot\infty=0$, in measure theory at least. $\endgroup$
    – user123641
    Dec 11 '17 at 13:57

If $f:S\to[0,\infty]$ is integrable, then $f \ge 0$ , thus $f=f^{+}$ and $f^{-}=0$.

Let $A:=\{x \in S: f(x)= + \infty\}$. For each natural $n$ we have $n \cdot 1_ A \le f$, hence

$n \mu(A)=\int_S n \cdot 1_ A d \mu \le \int_s f d \mu < \infty$ for all $n$. This gives $\mu(A)=0$.

Your second question: consider $f=1_{\mathbb Q}$. $f$ is Lebesque-integrable over $[0,1]$, but not Riemann- integrable over $[0,1]$

  • $\begingroup$ Thanks for your reply! How do you know that $n\mu(A) = \int_S 1_A d\mu$? What if there are infinitely many $x\in A$? $\endgroup$
    – titusAdam
    Dec 11 '17 at 15:14
  • $\begingroup$ Sorry, a typo ! It should read: $n \mu(A)=\int_S n \cdot 1_ A d \mu $. $\endgroup$
    – Fred
    Dec 12 '17 at 6:24

I only answer the first question of yours. The second question is asked here several times and you can find many interesting answers.

Okay, let's start. "Almost everywhere" means that a property holds everywhere except for a null set. What is a null set? A null set is a measurable set such that it's measure is zero. So you can for example say the following function is $1$ almost everywhere (using the Lebesgue measure) , where the function is $1$ in $\mathbb{R}\setminus\mathbb{Q}$ and something else in $\mathbb{Q}$. Because $\mathbb{Q}$ has Lebesgue measure zero (why?).

For the excercise you have a function $f$ that is non-negative, so $f^+=f$. To prove the statement $f<\infty$ almost everywhere, we do it by contradiction. Assume there is some non null set $K$ for which $f(x)=\infty$ for all $x\in K$. Then we have: \begin{align} \infty= \int_K fd\mu= \int_K f^+d\mu\leq \int_S f^+d\mu<\infty \end{align} A contradiction. Hence the claim follows.


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