0

I have some doubts regarding a question in a Lebesgue theory book by Spiegel.

The question is as follows:

$\begin{align} &\text{Given}\\ &\quad\text{a) } f_n(x)=e^{-nx^2} \text{ in } \;0\leq{x}\leq{1}\;\text{ and}\\ &\quad\text{b) } F(x)=\lim\limits_{n\to\infty}f_n(x)\;,\\ &\text{then what are the Baire functions classes that a)}\\ &\text{and b) belong to ?} \end{align}$

Thanks in advance, greetings.

Angelo
  • 12,328
Atha
  • 9

1 Answers1

0

Do you understand why $F(x)$ is discontinuous? Compare $F(0)$ to $F(x)$ for $x\neq 0$. This proves $F(x)$ is Baire class 1, at least according to the Wikipedia definition.

John Hopfensperger
  • 496
  • thank you!! to be honest, it seems to me to be the case that functions a) are continuous functions; on the other hand, given that the limit of a sequence of continuous function be a discontinuous function, would yield the result: that is, functions a) to be Baire class 0 while functions b) to be Baire class 1. – Atha Oct 16 '20 at 04:25
  • That's my understanding too. – John Hopfensperger Oct 16 '20 at 04:29
  • ps: Beside that, i wonder whether the discontinuity of F(x) could be brought about by means of some interval relation like $\lvert x-x_0 \rvert \leq{ε} \longrightarrow{\lvert F(x)-F(x_0) \rvert }\leq{ξ}$ greetings – Atha Oct 16 '20 at 04:42
  • No matter how small you take $\delta > 0$, $|x-0| < \delta$ fails to imply $|F(x) - F(0)| < 1$. – John Hopfensperger Oct 16 '20 at 04:48
  • thanks a lot!! i'll think about it and probably make some additional comment later on. greetings – Atha Oct 16 '20 at 04:54
  • could it be the following case: just as $x$ departs from $0$, $F(x)$ value "collapse" from its value at $0$, that is, from 1? if so, the function $F(x)$ would have a discontinuity point at $0$; namely, a "skip" type of discontinuity?; perhaps the notion of "nowhere dense set" could be called upon to explain something about such a discontinuity? thanks! – Atha Nov 11 '20 at 06:25