Show that $f<\infty$ almost anywhere. 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.
Questions:


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*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!
 A: 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 Lebesgue integrable over $[0,1]$, but not Riemann integrable over $[0,1]$
A: 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. 
