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Homework Statement
Given a measure space (\mathbb{R},<br /> \mathcal{B}(\mathbb{R}),\mu) define a function F_\mu : \mathbb{R}<br /> \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}<br /> F_\mu(x_n) = F_\mu(x) for any decreasing sequence \{ x_n<br /> \}^\infty_{n=1} \subset \mathbb{R} with limit \displaystyle<br /> \lim_{n \to \infty} x_n = x.)
Homework Equations
The Attempt at a Solution
\mathbb{F}_\mu non-decreasing means for any x_1,x_2 \in X suchthat x_1 < x_2 we have \mathbb{F}_\mu(x_1) \leq<br /> \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<br /> \mathbb{R}, x_1 > x_2 > \cdots > x_n > \cdots and hence
(-\infty, x_1] \supset (-\infty, x_2] \supset \cdots (-\infty, x_n]<br /> \supset \cdots. Also \displaystyle (-\infty, x] =<br /> \bigcap^\infty_{n=1} (-\infty, x_n]. Hence according to theorem 7.1
(5) we have \mu((-\infty,x_n]) \xrightarrow[k]{\infty}<br /> \mu((-\infty,x]) which is \displaystyle \lim_{n \to \infty}<br /> F_\mu(x_n) = F_\mu(x).
For any decreasing sequence \{x_n \}^\infty_{n=1} \subset<br /> \mathbb{R} such that \displaystyle \lim_{n \to \infty} x_n = -<br /> \infty, \lim_{n \to \infty} F_\mu(x_n) = F_\mu(x) =<br /> \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?