# Limit of functions in measure space

1. May 24, 2010

### complexnumber

1. The problem statement, all variables and given/known data

Given a measure space $$(\mathbb{R}, \mathcal{B}(\mathbb{R}),\mu)$$ define a function $$F_\mu : \mathbb{R} \to \mathbb{R}$$ by $$F_\mu(x) := \mu( (-\infty,x] )$$. Prove
that $$F_\mu$$ is non-decreasing, right-continuous and satisfies
$$\displaystyle \lim_{x \to -\infty} F_\mu(x) = 0$$. (Right continuous
here may be taken to mean that $$\displaystyle \lim_{n \to \infty} F_\mu(x_n) = F_\mu(x)$$ for any decreasing sequence $$\{ x_n \}^\infty_{n=1} \subset \mathbb{R}$$ with limit $$\displaystyle \lim_{n \to \infty} x_n = x$$.)

2. Relevant equations

3. The attempt at a solution

$$\mathbb{F}_\mu$$ non-decreasing means for any $$x_1,x_2 \in X$$ such
that $$x_1 < x_2$$ we have $$\mathbb{F}_\mu(x_1) \leq \mathbb{F}_\mu(x_2)$$. Since $$(-\infty,x_1] \subset (-\infty,x_2]$$,
theorem 7.1 (3) says $$\mu((-\infty,x_1]) \leq \mu(-\infty,x_2])$$,
which is $$\mathbb{F}_\mu(x_1) \leq \mathbb{F}_\mu(x_2)$$.

For any decreasing sequence $$\{x_n \}^\infty_{n=1} \subset \mathbb{R}$$, $$x_1 > x_2 > \cdots > x_n > \cdots$$ and hence
$$(-\infty, x_1] \supset (-\infty, x_2] \supset \cdots (-\infty, x_n] \supset \cdots$$. Also $$\displaystyle (-\infty, x] = \bigcap^\infty_{n=1} (-\infty, x_n]$$. Hence according to theorem 7.1
(5) we have $$\mu((-\infty,x_n]) \xrightarrow[k]{\infty} \mu((-\infty,x])$$ which is $$\displaystyle \lim_{n \to \infty} F_\mu(x_n) = F_\mu(x)$$.

For any decreasing sequence $$\{x_n \}^\infty_{n=1} \subset \mathbb{R}$$ such that $$\displaystyle \lim_{n \to \infty} x_n = - \infty$$, $$\lim_{n \to \infty} F_\mu(x_n) = F_\mu(x) = \mu((-\infty,-\infty]) = 0$$.

Are these correct answers? They don't even look like proofs, especially the third one. What should the proofs be like?

2. May 25, 2010

### Office_Shredder

Staff Emeritus
F is a function on the real numbers, so taking $$F_{\mu} (- \infty)$$ doesn't make any sense

3. May 25, 2010

### complexnumber

I think the question must be assuming we are using the extended real line. In my lecture notes the $$\infty$$ measure is allowed. Is my answers correct assuming extended real line?

4. May 26, 2010

### Office_Shredder

Staff Emeritus
While the measure might give values on the extended real line, the interval $$(-\infty, -\infty]$$ doesn't make any sense because on the one hand it doesn't contain $$- \infty$$, but on the other hand it does.

You should have a standard definition for what the limit of a function is as the value goes to plus or minus infinity involving the function getting arbitrarily close to the limit value as the input grows in magnitude (with positive or negative values)