Functions having the same integral are equal

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SUMMARY

If two functions \( f(x,y,z) \) and \( g(x,y,z) \) yield the same integral over any solid region \( D \), it is established that they are equal almost everywhere. This conclusion is grounded in measure theory, which defines the concept of equality in terms of the measure of the set where the functions differ. The discussion highlights the importance of understanding this principle for applications such as equating integral and differential forms in Gauss's law.

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  • Measure theory concepts, specifically "almost everywhere" equality
  • Understanding of triple integrals in multivariable calculus
  • Familiarity with Gauss's law in electromagnetism
  • Basic knowledge of function properties and continuity
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  • Study the concept of "almost everywhere" in measure theory
  • Explore the properties of triple integrals in multivariable calculus
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Suppose that for any solid region D, it is true that
[tex]\int\int\int_{D}f(x,y,z)dV = \int\int\int_{D}g(x,y,z)dV[/tex]

Then is it the case that f(x,y,z) is g(x,y,z). I am not sure if it's true but I seem to need it to equate the integral and differential forms of Gauss's law.

Any thoughts?

BiP
 
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Hint: consider f-g.
 
To be more precise: it is true almost everywhere.

If you have learned measure theory, then the words "almost everywhere" have a very precisely defined meaning. Otherwise, you can just take it very loosely, and think of examples like
$$f(x) = x^2; g(x) = \begin{cases} x^2 & \text{if } x \neq 0 \\ -1 & \text{ if } x = 0 \end{cases}$$
 

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