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Homework Statement
Given a measure space [tex](\mathbb{R},<br /> \mathcal{B}(\mathbb{R}),\mu)[/tex] define a function [tex]F_\mu : \mathbb{R}<br /> \to \mathbb{R}[/tex] by [tex]F_\mu(x) := \mu( (-\infty,x] )[/tex]. Prove
that [tex]F_\mu[/tex] is non-decreasing, right-continuous and satisfies
[tex]\displaystyle \lim_{x \to -\infty} F_\mu(x) = 0[/tex]. (Right continuous
here may be taken to mean that [tex]\displaystyle \lim_{n \to \infty}<br /> F_\mu(x_n) = F_\mu(x)[/tex] for any decreasing sequence [tex]\{ x_n<br /> \}^\infty_{n=1} \subset \mathbb{R}[/tex] with limit [tex]\displaystyle<br /> \lim_{n \to \infty} x_n = x[/tex].)
Homework Equations
The Attempt at a Solution
[tex]\mathbb{F}_\mu[/tex] non-decreasing means for any [tex]x_1,x_2 \in X[/tex] suchthat [tex]x_1 < x_2[/tex] we have [tex]\mathbb{F}_\mu(x_1) \leq<br /> \mathbb{F}_\mu(x_2)[/tex]. Since [tex](-\infty,x_1] \subset (-\infty,x_2][/tex],
theorem 7.1 (3) says [tex]\mu((-\infty,x_1]) \leq \mu(-\infty,x_2])[/tex],
which is [tex]\mathbb{F}_\mu(x_1) \leq \mathbb{F}_\mu(x_2)[/tex].
For any decreasing sequence [tex]\{x_n \}^\infty_{n=1} \subset<br /> \mathbb{R}[/tex], [tex]x_1 > x_2 > \cdots > x_n > \cdots[/tex] and hence
[tex](-\infty, x_1] \supset (-\infty, x_2] \supset \cdots (-\infty, x_n]<br /> \supset \cdots[/tex]. Also [tex]\displaystyle (-\infty, x] =<br /> \bigcap^\infty_{n=1} (-\infty, x_n][/tex]. Hence according to theorem 7.1
(5) we have [tex]\mu((-\infty,x_n]) \xrightarrow[k]{\infty}<br /> \mu((-\infty,x])[/tex] which is [tex]\displaystyle \lim_{n \to \infty}<br /> F_\mu(x_n) = F_\mu(x)[/tex].
For any decreasing sequence [tex]\{x_n \}^\infty_{n=1} \subset<br /> \mathbb{R}[/tex] such that [tex]\displaystyle \lim_{n \to \infty} x_n = -<br /> \infty[/tex], [tex]\lim_{n \to \infty} F_\mu(x_n) = F_\mu(x) =<br /> \mu((-\infty,-\infty]) = 0[/tex].
Are these correct answers? They don't even look like proofs, especially the third one. What should the proofs be like?