(New page: 2a. <math>( \Rightarrow )</math> Say <math>f</math> is A.C. Then <math>f</math> is of bounded variation, and since <math>f</math> is clearly nondecreasing, <math>f</math> must be bounded...) |
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<math>( \Leftarrow )</math> | <math>( \Leftarrow )</math> | ||
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| + | Now say <math>\sum_{n=1}^\infty m(G_n)<\infty</math>. | ||
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| + | I claim <math>f'(x) = \sum_{n=1}^\infty \chi_{G_n}(x)</math> a.e. as follows: | ||
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| + | For <math>h>0</math>, <math>\frac{f(x+h)-f(x-h)}{2h}=\frac{1}{2h}\sum_{n=1}^\infty m([x-h,x+h]\cap G_n)=\frac{1}{2h}\sum_{n=1}^\infty \int_{x-h}^{x+h} \chi_{G_n}=\frac{1}{2h} \int_{x-h}^{x+h} \sum_{n=1}^\infty \chi_{G_n}(x)</math> (MCT) | ||
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| + | Taking <math>lim_{h\rightarrow 0}</math> of both sides we get <math>f'(x)</math> on the left and <math>\sum_{n=1}^\infty \chi_{G_n}(x)</math> a.e. on the right (Lebesgue's Differentiation Theorem). | ||
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| + | Now, since <math>\sum_{n=1}^\infty m(G_n)<\infty</math>, we have <math>\int_0^1 f' = \int_0^1 \sum_{n=1}^\infty \chi_{G_n}= \sum_{n=1}^\infty \int_0^1 \chi_{G_n}=\sum_{n=1}^\infty m(G_n)<\infty</math> <math>\Rightarrow f'</math> is integrable on <math>[0,1]</math>. | ||
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| + | Finally, <math>f(x)-f(0)=\int_0^x f'</math>. | ||
| + | |||
| + | So <math>f</math> is A.C. | ||
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| + | b) | ||
Revision as of 12:46, 8 July 2008
2a.
$ ( \Rightarrow ) $ Say $ f $ is A.C. Then $ f $ is of bounded variation, and since $ f $ is clearly nondecreasing, $ f $ must be bounded.
In particular, $ \infty > f(1)=\sum_{n=1}^\infty m([0,1]\cap G_n) \Rightarrow \sum_{n=1}^\infty m(G_n)<\infty $.
$ ( \Leftarrow ) $
Now say $ \sum_{n=1}^\infty m(G_n)<\infty $.
I claim $ f'(x) = \sum_{n=1}^\infty \chi_{G_n}(x) $ a.e. as follows:
For $ h>0 $, $ \frac{f(x+h)-f(x-h)}{2h}=\frac{1}{2h}\sum_{n=1}^\infty m([x-h,x+h]\cap G_n)=\frac{1}{2h}\sum_{n=1}^\infty \int_{x-h}^{x+h} \chi_{G_n}=\frac{1}{2h} \int_{x-h}^{x+h} \sum_{n=1}^\infty \chi_{G_n}(x) $ (MCT)
Taking $ lim_{h\rightarrow 0} $ of both sides we get $ f'(x) $ on the left and $ \sum_{n=1}^\infty \chi_{G_n}(x) $ a.e. on the right (Lebesgue's Differentiation Theorem).
Now, since $ \sum_{n=1}^\infty m(G_n)<\infty $, we have $ \int_0^1 f' = \int_0^1 \sum_{n=1}^\infty \chi_{G_n}= \sum_{n=1}^\infty \int_0^1 \chi_{G_n}=\sum_{n=1}^\infty m(G_n)<\infty $ $ \Rightarrow f' $ is integrable on $ [0,1] $.
Finally, $ f(x)-f(0)=\int_0^x f' $.
So $ f $ is A.C.
b)
