Re: Analysis with integral.

" My question is....

If f(x) is a bounded and continuous on [a,b]
and f(x) = g(x) at except countably discontinuous points,
(Namely, g(x) = f(x) , (x not x_1, x_2, x_3, ....) and
g(x) is not continuous at x_1, x_2, x_3, .....)

Then int_{a to b} f(x) dx = int{a to b} g(x) dx.

is this possible ?"

It is true, if that is what you mean:

Informally, if you have only countably many non-zero
points, you can make the partition width small-enough
so that the sum becomes 0 .


consider

h(x)= g(x)-f(x)=0 except at x_1,...,x_n,....

Consider a Riemann sum in which the values

x_i* , i.e, the values in the i-th element of the

partition that you select for :


Sum (n=1,..,oo)f(x_i*)dx_i


Let M=maxf(x) over [a,b] (assume wolg that f>=0)


Them above sum is bounded above by:

Sum(n=1,...,oo)Mdx_i =


M[ Sum(n=1,...,oo)dx_i]<=M(b-a)

Now make the partition width |P|=maxdx_i

small-enough, and the Riemann sum is zero.

()
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