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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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You're right. It looks like Natanson didn't realize this -- after "the function $[f(x)]_N$ (for arbitrary $N$) is equivalent to zero", it would have sufficed to just state that it differs from zero precisely where $f$ differs from zero to conclude that $f$ is equivalent to zero, dispensing with the rest of the argument. –  joriki Apr 15 '11 at 19:39
    
@joriki: could it be? was thinking that I might have missed something here, given that Natanson is well-known in this field, and also the author of his numerous nice-written volumes. –  Qiang Li Apr 15 '11 at 19:55
    
The numerosity of the volumes might actually be an argument in favour of thinking he might not have checked each and every argument meticulously :-) –  joriki Apr 15 '11 at 20:25

1 Answer 1

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 $A$.

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