Is g integrable if it equals f at all but a finite number of points?

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In the discussion, it is established that if a function g equals an integrable function f at all but a finite number of points within a rectangle A, then g is also integrable, and the integrals of both functions over A are equal, i.e., ∫A f = ∫A g. The discussion specifically references the Riemann integral and emphasizes the importance of analyzing the impact of points of inequality on upper and lower integrals. The conclusion is that the integrability of g is maintained despite the finite discrepancies with f.

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i don't know where to start on this problem. could someone help me please? thanks.

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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What about g - f?
 
jeanf said:
i don't know where to start on this problem. could someone help me please? thanks.

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]
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.
 

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