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I am reading Natanson's "Theory of functions of a real variable". In Theorem 6 of $\S 1$ of Chapter VI when he is discussing summable functions, I have a small question. I re-phrase it as below.

Let $f$ be a non-negative function defined on an interval $[a,b]$. Let $N$ be a natural number. Define

$[f(x)]_N=\begin{cases} f(x) \text{ if } f(x) \le N, \\ \newline N \text{ if } f(x) \gt N. \end{cases}$

Then we easily know $\lim_{N\to +\infty}{[f(x)]_N}=f(x)$ pointwise on $[a,b]$.

My question is:

  • Is it true that $A_N=\{x:[f(x)]_N\ne 0\}$ the same for all natural number $N$? I would think so. Since these sets are all equal to $A=\{x:f(x)\ne 0\}.$

If so, then there is no point to write $A=\bigcup_{N=1}^\infty{A_N}$ because they are all equal to $A$, correct?

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    The numerosity of the volumes might actually be an argument in favour of thinking he might not have checked each and every argument meticulously :-)2011-04-15

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Correct. As long as we aren't including 0 as a natural number (which some people do), then $[f(x)]_N=0$ if and only if $f(x)=0$ for every natural number $N$, so every $A_N$ is equal to $A$.