(New page: In Progress... Ben <math>g(x)</math> is monotone: Take <math>x<y</math>. Then <math>g(x) = \sum{f_n(x)} \le \sum{f_n(y)} = g(y)</math>. Since <math>f_n(x) \in AC[0,1]</math> and <math>f_...) |
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| − | + | Since <math>f_n(x) \in AC[0,1]</math> and <math>f_n(0)=0, f(x) = \int_0^x{f_n'(t)dt} </math> | |
| − | + | <math>g(x) = \sum_{n=0}^\infty{\int_0^x{f_n'(t)dt}}= lim_{N\rightarrow\infty}\sum_{n=0}^N{\int_0^x{f_n'(t)dt}} = lim_{N\rightarrow\infty}\int_0^x{\sum_{n=0}^N{f_n'(t)dt}}</math> becuase the sum is finite. | |
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| + | Since <math>f_n(x)</math> is monotone, <math>f_n'(x) \ge 0</math>. Therefore, <math>\sum_{n=0}^N{f_n'(t)dt}</math> is monotone as a function of <math>N</math>. | ||
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| + | <math>g(x) \stackrel{\rm MCT} {=} \int_0^x{lim_{N\rightarrow\infty}\sum_{n=0}^N{f_n'(t)}dt} = \int_0^x{\sum_{n=0}^\infty{f_n'(t)}dt}</math> | ||
| + | Call <math>h(x) = \sum_{n=0}^\infty{f_n'(t)}</math>. <math>\int_0^1 |h(t)|dt = g(1) < \infty</math> so <math>h(x)\in L^1[0,1]</math> | ||
| − | + | <math>g(x) = \int_0^x h(t) dt</math> so <math>g(x) \in AC[0,1]</math> by the absolute continuity of the integral. | |
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| − | + | --[[User:Bbartle|Bbartle]] 11:43, 10 July 2008 (EDT) | |
Latest revision as of 11:43, 10 July 2008
Since $ f_n(x) \in AC[0,1] $ and $ f_n(0)=0, f(x) = \int_0^x{f_n'(t)dt} $
$ g(x) = \sum_{n=0}^\infty{\int_0^x{f_n'(t)dt}}= lim_{N\rightarrow\infty}\sum_{n=0}^N{\int_0^x{f_n'(t)dt}} = lim_{N\rightarrow\infty}\int_0^x{\sum_{n=0}^N{f_n'(t)dt}} $ becuase the sum is finite.
Since $ f_n(x) $ is monotone, $ f_n'(x) \ge 0 $. Therefore, $ \sum_{n=0}^N{f_n'(t)dt} $ is monotone as a function of $ N $.
$ g(x) \stackrel{\rm MCT} {=} \int_0^x{lim_{N\rightarrow\infty}\sum_{n=0}^N{f_n'(t)}dt} = \int_0^x{\sum_{n=0}^\infty{f_n'(t)}dt} $
Call $ h(x) = \sum_{n=0}^\infty{f_n'(t)} $. $ \int_0^1 |h(t)|dt = g(1) < \infty $ so $ h(x)\in L^1[0,1] $
$ g(x) = \int_0^x h(t) dt $ so $ g(x) \in AC[0,1] $ by the absolute continuity of the integral.
--Bbartle 11:43, 10 July 2008 (EDT)
