Equality of integrals => equality of integrands

  • Thread starter Derivator
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In summary, the argument for the equality of integrals for any domain A is justified by considering the function f(x) - g(x) and using Tschebyshev's inequality to show that the set where f is greater than g has measure zero. This applies to both continuous and discontinuous functions, with the exception being a set of measure zero.
  • #1
Derivator
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hi folks,

one often reads

[itex]\int_A f(x) dx = \int_A g(x) dx [/itex] for arbirary A, thus f(x) = g(x), since the equaltiy of the Integrals holds for any domain A.

I don't see, why the argument "...for any domain A..." really justifies this conclusion.

Can someone explain this to me, please?
 
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  • #2
Derivator said:
hi folks,

one often reads

[itex]\int_A f(x) dx = \int_A g(x) dx [/itex] for arbirary A, thus f(x) = g(x), since the equaltiy of the Integrals holds for any domain A.

I don't see, why the argument "...for any domain A..." really justifies this conclusion.

Can someone explain this to me, please?

For continuous functions and argument might go like this. If f and g aren't identical, there is a point a where one of them, say f(a), is greater than g(a). So there is an interval I containing a where f(x) > g(x). But then ##\int_I f(x)-g(x)\, dx > 0## contradicting the assumption "for any domain A".
 
  • #3
And for discontinuous functions, the result might not even be true! It needs to be:

If [itex]\int_A f(x)dx = \int_A g(x)dx[/itex] then [itex]f=g[/itex] almost everywhere.
 
  • #4
For the more general case the conclusion would be f(x)=g(x), except on a set of measure 0.
 
  • #5
I think you can do something like the following. Assuming the integrals of f and g are equal for every set A (and f and g are obviously measurable):

Consider the function [itex] f(x) - g(x) [/itex]. Let E be the set of x's in A where this function is positive. Define

[itex] E_{1/n} = \left{ x \in E : f(x) - g(x) > 1/n \right} [/itex]

Thus by Tschebyshev,

[itex] m(E_{1/n}) \leq n \int_{ E_{1/n} } f - g = 0 [/itex]

Therefore

[itex] m(E) = \cup_{n=1}^\infty m(E_{1/n}) = 0 [/itex]

ie, the set where f is strictly greater than g has measure zero.

You can do basically the same argument to show that the set where g is strictly greater than f has measure zero. So f and g are the same for ae x. It's not really a for any domain A type question, it just needs to be the case for these E_{1/n}'s.
 
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1. What is the meaning of "equality of integrals => equality of integrands"?

This statement means that if two integrals have the same value, then the integrands (the functions being integrated) must also be equal.

2. Why is equality of integrals important?

Equality of integrals is important because it allows us to solve integration problems more efficiently. By knowing that two integrals with the same value have equal integrands, we can simplify the problem by only having to find one integral.

3. Is equality of integrals always true?

Yes, equality of integrals is a fundamental property of integration and is always true.

4. Can equality of integrals be used to solve all integration problems?

No, equality of integrals is only one aspect of solving integration problems. Other techniques, such as substitution and integration by parts, may also be needed to solve certain integrals.

5. How can we prove equality of integrals => equality of integrands?

This property can be proven using the Fundamental Theorem of Calculus, which states that if a function is continuous, then the integral of its derivative is equal to the original function. By using this theorem, we can show that if two integrals have the same value, then their integrands must also be equal.

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