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I am trying to work through an exercise (on my own) out of Resnick's A Probability Path.

One question states the following:

Suppose $-\infty<a\le b<\infty$ and assume we have an indicator function of the form $1_{(a,b]}(x)$. Can this function be approximated by bounded and continuous functions.

Said another way, does there exist a sequence $f_n: 0\le f_n\le 1$ such that $f_n\rightarrow1_{(a,b]}$ pointwise.

Using the hint in the text, my attempted solution was:

Attempt at a solution:

$f_n$ such that: $f_n(x) = \begin{cases} 0, & x \leq a, x \geq b + \frac1n \\ 1, & a + \frac1n < x \leq b \end{cases}$ and linear otherwise

The problem (or one problem?) is that this doesnt appear continuous to me. I could also use some help in learning how to be more explicit about limits. For example, why does $x>b$ imply $f_n\rightarrow 0$. I "know" this intuitively, but obviously need help explaining it with precision.

Thanks as always!

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up vote 2 down vote accepted

You are looking at functions which are constant ($0$ or $1$) everywhere with the exception of the intervals $(a, a+1/n)$ and $(b, b+1/n)$. In these intervals you have linear functions with increasing slope which grow from $0$ to $1$ or the other way round.

To prove pointwise (!) convergence just look at:

i) $x\le a$. There your sequence is constantly zero, so it converges to $0$

ii) any $x \in (a, b]$. Choose $N$ such that $a+1/n < x$ for $n\ge N$ . Then $f_n(x) = 1$ for each such $n$, hence $f_n(x)\rightarrow 1$

iii) $x>b$. Choose $N$ such that $x\le b+1/n$ for $n\ge N$. Then $f_n(x)=0$ hence $f_n(x)\rightarrow 0$.

Then you are done.

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thank you for the notes! – Justin Jun 16 '12 at 0:23

Try piecing $1_{(a+ \frac{1}{n}, b ]}$ together with a positively-sloped line segment from $a$ to $a+\frac{1}{n}.$ Let me know if you'd like more detail.

EDIT: In other words, your sequence should look something like this

$f_n(x) = \begin{cases} 0 \textrm{ if } x < a \textrm{ or } x >b \\ nx - na \textrm{ if } a\le x \le a + \frac{1}{n}\\ 1 \textrm{ if } a + \frac{1}{n} < x \le b \end{cases}$

Try drawing a picture to get a sense for how the line segment was constructed. To show that this converges to pointwise to $1_{(a,b]}$, you should consider the cases of $x\notin (a,b]$ and $x\in (a,b]$ separately. The first case is pretty easy - the second case doesn't require too much work if you are familiar with Archimedes' Property.

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thank you for the response. Detail would be helpful as I am learning this from scratch. What do you mean exactly by "piece together"? Are you saying to construct a rectangle for the approximation? – Justin Jun 15 '12 at 19:33
@Justin, let me know if the edit above can be clarified any more. – user17794 Jun 15 '12 at 19:44
Thank you Tim. I think it makes sense now. – Justin Jun 16 '12 at 0:22

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