Hölder inequality conditions for $L_p$ spaces? The Hölder inequality is the statement that if $f,g$ are measurable functions then 
$$ \|fg \|_1 \le \|f\|_p \|g\|_q$$
if $p,q$ are such that ${1\over p}+ {1 \over q} =1$.
But it's not clear to me why, if both $\|f\|_p$ and $ \|g\|_q$ are finite, this implies that $\|fg\|_1$ is, too. 

Is there an assumption missing in the statement on Wikipedia? (Like e.g. $f \in L_1 \cap L_p$.) Is there really no example of $f,g$ measurable such that $\|f\|_p$ and $ \|g\|_q$
  are finite but $\|fg\|_1 = \infty$?

I know that if the measure space $X$ is finite then there is an $L_p$ space inclusion $L_q \subseteq L_p$ if $p < q$. But finiteness is not in the assumptions of the Hölder inequality. 
 A: This is part of the point of Hölder's inequality. It shows that if $f\in L^p$, $g\in L^q$ and $\frac1p+\frac1q=1$ then $fg\in L^1$. To see "why" this holds, I suggest you examine a proof of Hölder's inequality. I can provide one if you do not have one.
EDIT: The proof relies on Young's inequality: if $x,y\ge0$ then $xy\le\frac{x^p}p+\frac{y^q}q$. This inequality is obvious if $x$ or $y$ are zero, so assume $x,y>0$. For any $t\in[0,1]$, we have
$$\log(tx+(1-t)y)\ge t\log(x)+(1-t)\log(y)$$
which can be seen by observing the graph of $\log$. Setting $t=\frac1p$, we get
$$\log\left(\frac{x^p}p+\frac{y^q}q\right)\ge \frac1p\log(x^p)+\frac1q\log(y^q)=\log(xy)$$
from which Young's inequality follows.
Now we prove Hölder's inequality. If $f$ or $g$ are zero then it is trivial, so assume $\|f\|_p,\|g\|_q>0$. Define $F=\frac f{\|f\|_p},G=\frac g{\|g\|_q}$. Then $F\in L^p,G\in L^q$ with $\|F\|_p=\|G\|_q=1$. Using Young's inequality, we have
$$\int_X|FG|\ d\mu\le\int_X\left(\frac{|F|^p}p+\frac{|G|^q}q\right)\ d\mu=\frac1p\|F\|_p^p+\frac1q\|G\|_q^q=1.$$
We have shown
$$\frac1{\|f\|_p\|g\|_q}\int_X|fg|\ d\mu=\int_X\left|\frac f{\|f\|_p}\frac g{\|g\|_q}\right|\ d\mu\le1,$$
so we are done. Note that by carefully keeping track of where equality occurs in Young's inequality, one can show that equality occurs if and only if $|f|^p$ and $|g|^q$ are linearly dependent as elements of $L^1$.
A: 
not clear to me why, if both $\|f\|_p$ and $\|g\|_q$ are finite, this implies that $\|fg\|_1$ is, too. 

Note that
$$ 1 < \infty $$
Therefore,
$$ \text{if $b=1$ and $a=\infty$ then $b < a$.} $$
Contraposing,
$$ \text{if $a\le b$ then it is not the case that $b=1$ and $a=\infty$.} $$
Now, Hölder's inequality assures us that if $a=\|fg\|_1$ and $b=\|f\|_p\|g\|_q$ then $a\le b$.  Therefore it is not the case that $\|fg\|_1=\infty$ and $\|f\|_p\|g\|_q = 1$.
