In Rudin, while defining the concept of 'pointwise bounded', it says: if there exists a finite-valuded function $\phi$ defined on $E$ such that $|f_{n}(x)|<\phi (x)$. Here, I am quite puzzled by the definition of finite-valued. Is $f(x)=1/x$ an example of finite-valued function, since when $x$ tends to zero, it tends to infinity, how could it be classified as finite-valued? And any example of a function which is not finite-valuded? Thanks so much for your help!
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3 Answers
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Finite-valued means that a function only takes values in the real line (i.e. $(-\infty,+\infty)$). $f(x)=1/x$ is finite-valued since $1/x\in(-\infty,+\infty)$ for each $x\ne0$. A function that is not finite-valued takes values in the extended real line (i.e. $[-\infty,+\infty]$). For example, the Lebesgue measure on the real line ($\lambda((-\infty,+\infty))=+\infty$). Finite-valued does not mean bounded, $f(x)=1/x$ is not bounded, but it is finite-valued.
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Earlier in Rudin he defines the extended real numbers $R\cup\{\pm \infty\}$. He means that the function is real valued and not extended-real valued.
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$f(x) = 1/x$ is finite valued if you define it in $\mathbb R \setminus \{0\}$. If you define it in $[0,+\infty)$ by letting $f(0)=+\infty$, it is not finite valued.
answered Mar 13, 2014 at 8:18