Proof of Hölder's inequality for improper integrals 
If $f,g\in \mathscr{R} (\alpha)$ then Hölder's inequality is also true for improper integrals.


I can't understand how they get the last equality. We know that $\int \limits_{0}^{1}f\,dx$ exists i.e. exists $\lim \limits_{c\to 0}\int \limits_{c}^{1}f\,dx$. How they conclude that exists also $\int \limits_{0}^{1}|f|^p\,dx$.
Can anyone explain it please?
 A: This is an exercise in Rudin's Principle of Mathematical Analysis (Chapter 6 Problem 10c). 
One has the assumptions that $p$ and $q$ are positive real numbers such that
$$
\frac1p+\frac1q=1.
$$
Also, $f$ and $g$ should be complex functions in $\mathscr{R}(\alpha)$. More precisely,  $f$ and $g$ are Riemann-Stieltjes integrable with respect to some $\alpha$ over $(c,1]$ for any $c\in(0,1)$.
For any $c\in(0,1)$, one has (this is Chapter 6 Problem 10b)
$$
\left|\int_c^1fg\ d\alpha\right|\leq \left\{\int_c^1|f|^p\ d\alpha\right\}^{1/p}
\left\{\int_c^1|g|^q \ d\alpha\right\}^{1/q}.
$$
If 
$$
\lim_{c\to 0+}\left\{\int_c^1|f|^p\ d\alpha\right\}^{1/p}=\infty
$$
then
$$
\left|\int_0^1fg\ d\alpha\right|\leq \left\{\int_0^1|f|^p\ d\alpha\right\}^{1/p}
\left\{\int_0^1|g|^q \ d\alpha\right\}^{1/q} 
$$
is automatically true. Likewise for 
$$
\lim_{c\to 0+}\left\{\int_c^1|g|^q\ d\alpha\right\}^{1/q}=\infty.
$$
When 
$$
\lim_{c\to 0+}\left\{\int_c^1|f|^p\ d\alpha\right\}^{1/p}<\infty
$$
and 
$$
\lim_{c\to 0+}\left\{\int_c^1|g|^q\ d\alpha\right\}^{1/q}<\infty,
$$
you have the last equality in OP by definition of the improper integrals.
