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

In article <1115068833.500016.305340@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
<agapito6314@xxxxxxx> wrote:
>The integral (I) of function f with respect to measure u is defined by
>
>I (f du) = I (f+ du) - I (f- du)
>
>where f+ and f- are the positive and negative parts of f. Now if
>f = v - w, where u and v are any non-negative functions, my text
>(Bartle) goes on to say that "it is easy to see" that
>
>I (f du) = I (v du) - I (w du)
>
>How is it possible to arrive at this assuming only the definition of
>(Lebesque) integral of non-negative functions?

By definition you have

I (f du) = I (v-w)+ du - I (v-w)- du.

(v-w)+ is v-w when v>=w, and 0 elsewhere.
(v-w)- is w-v when v<=w, and 0 elsewhere.

Let A1 be the set where v>=w, A2 where v<=w. Note that on A1 intersect
A2, v=w, so v-w = 0.

Then

I (v-w)+du = I_{A1}(v-w) du
I (v-w)-du = I_{A_2}(w-v) du.

We can also write

I (v du) = I_{A_1} (v du) + I_{A_2} (v du) - I_{A_1\cap A_2} (v du)
I (w du) = I_{A_1} (w du) + I_{A_2} (w du) - I_{A_1\cap A_2} (w du)

so

I(v du) - I(w du) = I_{A_1}(v du) - I_{A_1} (w du)
+I_{A_2}(v du) - I_{A_2}(w du)

[the other two terms cancel, since v = w on A_1 \cap A_2

= I_{A_1}(v - w)du (since v>=w in A_1)
- I_{A_2}(w-v)du (since w>=v in A_2).

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...


--
======================================================================
"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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