- #1

jeanf

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let [tex] f: A -> R [/tex] be an integrable function, where A is a rectangle. If g = f at all but a finite number of points, show that g is integrable and [tex] \int_{A}f = \int_{A}g. [/tex]

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- Thread starter jeanf
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- #1

jeanf

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let [tex] f: A -> R [/tex] be an integrable function, where A is a rectangle. If g = f at all but a finite number of points, show that g is integrable and [tex] \int_{A}f = \int_{A}g. [/tex]

- #2

Hurkyl

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What about g - f?

- #3

lurflurf

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Do you mean the Reimann integral? For each value that they are not equal consider a small interval (small enough that only one point of inequality is included). Then |Sup(f)-Sup(g)|=|f(x*)-g(x*)|>0. Then consider the effect of all the points of inequallity on the upper integrals, then likewise for the lower integrals.jeanf said:

let [tex] f: A -> R [/tex] be an integrable function, where A is a rectangle. If g = f at all but a finite number of points, show that g is integrable and [tex] \int_{A}f = \int_{A}g. [/tex]

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