Boot20
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Is the integral over a set of measure zero always equals to zero? Can the integral be undefined?
The integral over a set of measure zero is definitively zero for ordinary functions, according to Lebesgue integration principles. This conclusion is supported by the convention that \(\infty \cdot 0 = 0\), which prevents inconsistencies when dealing with functions that may take on infinite values. Riemann integration, however, does not apply in cases involving extended-real-number-valued functions. The monotone convergence theorem further reinforces that if a function approximated by a sequence of functions \(f_n\) has an integral of zero, then the integral of the original function \(f\) must also be zero.
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mathman said:It may be undefined if the function itself is peculiar with infinity as its value. For ordinary functions the integral will be 0.
wayneckm said:But apparently to me \infty \cdot 0 = 0 should be adopted
Else, if f admit \infty on set A of measure 0, we may use f_{n} = n on A to approximate f from below, then, the integral of f_{n} is zero, by monotone convergence theorem, the integral of f should be zero as well. If we do not define \infty \cdot 0 = 0, we may get inconsistency in this case?
Boot20 said:But what if the f_{n} are defined over sets of measure noe intgen-zero, but that the sum of the measure of those sets converges to zero?