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In the book of Analysis on Manifolds at page 90 question 4, it is asked that

We say $f:[0,1] \to \mathbb{R}$ is increing if $f(x_1)<f(x_2)$ whenever $x_1<x_2$. If $f, g : (0, 1] \to \mathbb{R} $ are increing and non-negative, show that the function $h(x, y) = f(x)g(y)$ is integrable over $[0, 1]^2$.

and the same question has been asked in here; however, how do we know the values $f(1)$ and $g(1)$ are finite ? I mean there is nothing that assumes the functions are bounded in the statement of the question as far as I can see.

Our
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We have $f, g : (0, 1] \to \mathbb{R}$, hence $f((0,1]) \subseteq \mathbb R$ and $g((0,1]) \subseteq \mathbb R$ , therefore $f(1),g(1) \in \mathbb R$.

Fred
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    the domain is $I = [0,1]$, and not $(0,1]$. – Our Jun 15 '18 at 11:20
  • and what about $f(x) = 1/x$ ? isn't codomain of $f$ is $\mathbb{R} $ ? – Our Jun 15 '18 at 11:20
  • or $f(x) = 1/(1-x)$ ? – Our Jun 15 '18 at 11:21
  • @onurcanbektas the domain of $f(x)$ is $(0,1]$, so $f(1)$ is a real number (it is the definition of function). – Ixion Jun 15 '18 at 11:23
  • $f(x) = 1/(1-x)$ is not defined at $x=1$. @onurcanbektas –  Jun 15 '18 at 11:24
  • @JohnMa and Fred, then why do we have a concept called "bounded function" then ? After all, for $f(x) = 1/x$, where $f: (0,1]$, the value of $f$ for any $x$ in the domain has some value in $\mathbb{R}$ (hence not $\infty$). – Our Jun 15 '18 at 11:37
  • @JohnMa and Fred, Ok I have understood what I'm missing. Thanks for the help. – Our Jun 15 '18 at 11:42