Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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

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 $A$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.