Re: "It is easy to see...."

In article <1115225734.578032.86420@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
<agapito6314@xxxxxxx> wrote:
>
><snip>
>>
>> I am assuming, of course, that you know that if 0<= f <= g, then
>>
>> I (g-f)du = I gdu - I fdu, but that may not be a given. I used to
>know
>> all this stuff cold, but it's been too many years...
>
>Actually I don't know. Is there a simple proof of this?

Since 0<= f <= g, then both g-f and f are nonnegative. So

I (g du) = I( (g-f) + f)du = I(g-f)du + I fdu.

Therefore, I(g-f)du = I (gdu) - I (fdu).


--
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"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes")
======================================================================

Arturo Magidin
magidin@xxxxxxxxxxxxxxxxx

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