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It's a bulky four part question, most of which I've already worked through, but towards the end it begs for the construction of sets based on triangle functions and rectangular functions, which has so far strained my abilities. Here is the question:


Let $\{r_n\}$ be a dense subset of $\Bbb R$. Let $g_n:\Bbb R \rightarrow [0,1]$ be continuous $\forall n\in\Bbb N$. Assume $g_n(r_n)=1$ $\forall n$. Assume $\sum_{n\in\Bbb N} m(\{g_n>0\})<\infty$. Let $g=\sum_{n\in \Bbb N}g_n$.

  1. Show that $g$ is defined and finite almost everywhere.
  2. Show that $g$ is not bounded on any open interval (non-trivial)
  3. Let $g_n$ be a triangle function with height 1 and base $\frac 1 {2^n}$. Explicitly construct a set $F \subset [0,1]$ such that $m([0,1]\setminus F)<\frac 1 8$ and that $g_n \rightarrow 0$ on $F$ almost everywhere.
  4. Let $g_n$ be a rectangular function with height 1 and base $\frac 1 {2^n}$. Explicitly construct a set $F \subset \Bbb R$ such that $m(\Bbb R \setminus F)<\frac 1 8$ and a continuous function $f$ on $\Bbb R$ where $g=f$ on $F$.

So yes it's a long one. My work so far towards 1 and 2 are posted below. Being checked over would be great!

3 and 4 have stumped me pretty good, I am having trouble relating a triangle/rectangle function to the given conditions.


My Solutions in an answer

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2 Answers 2

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My Solutions

1 and 2. First off, assumptions:

  • Let $\{r_n\}$ be a dense subset of $\Bbb R$.
  • Let $g_n:\Bbb R \rightarrow [0,1]$ be continuous $\forall n\in\Bbb N$.
  • Assume $g_n(r_n)=1$ $\forall n$.
  • Assume $\sum_{n\in\Bbb N} m(\{g_n>0\})<\infty$.
  • Let $g=\sum_{n\in \Bbb N}g_n$.

Defined: Since $g_n$ maps to $[0,1]$, $g$ must be greater than or equal to zero since it is the sum of a bunch of positive/zero values. It will either converge to some number or diverge (to infinity). In either case it is defined since there is no alternating behavior. $\Box$

Finite AE: Since $\sum_{n\in\Bbb N} m(\{g_n>0\})< \infty$, there must be some point in the sequence for which the tail becomes arbitrarily small, for some $N \in \Bbb N$. I.e. when $k \ge N$ $\Rightarrow \sum_{n=k}^\infty m(\{g_n> 0\}) < \epsilon$.

Now notice $m(\bigcup_{n=N}^\infty\{g_n>0\}) \le \sum_{n=N}^\infty m(\{g_n>0\}) < \epsilon$. Now that we observe that $\bigcup_{n=N}^\infty\{g_n> 0\}$ is less than epsilon we can remove this from our consideration. So examining our function on the set $\Bbb R \setminus \bigcup_{N}^\infty\{g_n> 0\}$ we have that $g= \sum_{n=1}^\infty g_n= \sum_{n= 1}^{n— 1} g_n$. So $g$ is finite almost everywhere. (does this make sense?) $\Box$

Show not bounded on open interval: Consider some arbitrary open interval $I=(a,b)$. Inside $I$ exists some elements of $\{r_n\}$, say $r_i$ and $r_j$, where $a<r_i<r_j<b$ WLOG. Since $\{r_n\}$ is dense you can always find another $r_k$ in between, i.e. $a<r_i<r_k<r_j<b$ etc etc etc. You can continue to repeat this process to see that there are infinite number of $r_n$'s within $(a,b)$.

This implies that $\sum_{n=1}^\infty g_n$ is not bounded since $g_n(r_n)=1$ at each $r_n \in (a,b)$ and there are an infinite number of $r_n$'s within $(a,b)$ due to density illustrated above. (does this make sense?) $\Box$

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Hint for (3) and (4): you can in fact make all but finitely many $g_n$ equal to $0$ on $F$.

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By $E$ do you mean $F$? Also, I'm not quite sure what you mean, I was under the impression that $g_n$ had to be a triangle function defined for all the $r_n$ in the dense subset. –  KUSH Nov 20 '12 at 21:23
1  
Sorry, I did mean $F$. $g_n$ is a triangle function in (3), but the set where is is nonzero is disjoint from $F$. The fact that $r_n$ is dense is no problem, because $F$ is not an open set. –  Robert Israel Nov 21 '12 at 0:08

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