Re: a very simple proof

koolguyuf a écrit :
On Jun 18, 11:23 am, s...@xxxxxxxxxxxxxx wrote:
On 18 Jun, 16:08, koolguyuf <tdgosw...@xxxxxxxxx> wrote:

I need to convince myself of the following:
If \int f(x)_0^x = \int_0^x g(x),
then f(x) = g(x)
where both f(x) and g(x) are non-negative and the range of
integration is from 0 to x.
Your assertion is badly written, ambiguous and wrong. It would be
better written as
\int_0^x f(y) dy = \int_0^x g(y) dy.
The ambiguity is whether the equality is true for a given value of x
or for all x.

Counterexample 1)
Let x=1, f(y)=1 and g(y)=2y.
Then \int_0^x f(y) dy = 1 = \int_0^x g(y) dy
but f(y) and g(y) are different functions.

Counterexample 2)
Let f(y)=1, g(y)=1 for y irrational and g(y)=0 for y rational
Then \int_0^x f(y) dy = x = \int_0^x g(y) dy for all x
but f(y) and g(y) are different functions.

The best you can do is say that if for all x
\int_0^x f(y) dy = \int_0^x g(y) dy,
then f(y) = g(y) except on a set of measure zero.

Thanks forthe correction and the guidance. But can the following be
proved :

If for all x
\int_0^x f(y) dy = \int_0^x g(y) dy,
then f(y) = g(y) except on a set of measure zero.


Yes, it can be proved. But your following question leaves doubt if you know the *meaning* of this result (for instance what is "measure"...)




Moreover, is the converse also true. That is, if f(y)=g(y) for all y,
then is
\int_0^x f(y) dy = \int_0^x g(y) dy,

true for every x?

This is a consequence of the properties of equality. As soon as the symbol \foobar is defined , then if f(y)=g(y) for all y, \foobar f(y) = \foobar g(y)...




Thanks
TD


.

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