This shows you the differences between two versions of the page.
| Both sides previous revision Previous revision Next revision | Previous revision | ||
|
math105-s22:notes:lecture_12 [2022/02/24 07:58] pzhou [Pugh 6.8: Vitali Covering] |
math105-s22:notes:lecture_12 [2026/02/21 14:41] (current) |
||
|---|---|---|---|
| Line 32: | Line 32: | ||
| $$ \int F_+(x) dx = \int F_- (x) dx $$ | $$ \int F_+(x) dx = \int F_- (x) dx $$ | ||
| hence $F_+(x) = F_-(x)$ for almost all $x$. Thus, for a.e. $x$, we have $\uint f(x,y) dy = \lint f(x,y) dy$, thus $\int f(x,y) dy$ exists for a.e. x. | hence $F_+(x) = F_-(x)$ for almost all $x$. Thus, for a.e. $x$, we have $\uint f(x,y) dy = \lint f(x,y) dy$, thus $\int f(x,y) dy$ exists for a.e. x. | ||
| + | |||
| + | ==== A Lemma ==== | ||
| + | Suppose $A$ is measurable, and $B \In A$ any subset, with $B^c = A \RM B$. Then | ||
| + | $$ m(A) = m^*(B) + m_*(B^c) $$ | ||
| + | Proof: | ||
| + | $$ m^*(B) = \inf \{ m(C) \mid A \supset C \supset B, C \text{measurable} \} = \inf \{ m(A) - m(C^c) \mid A \supset C \supset B, C \text{measurable} \} $$ | ||
| + | $$ = m(A) - \sup \{ m(C^c) \mid A \supset C \supset B, C \text{measurable} \} = m(A) - \sup \{ m(C^c) \mid C^c \subset B^c, C^c \text{measurable} \} = m(A) - m_*(B^c) $$ | ||
| ===== Pugh 6.8: Vitali Covering ===== | ===== Pugh 6.8: Vitali Covering ===== | ||