How Do You Calculate the Lebesgue Integral of e^x from 1 to 10?

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SUMMARY

The Lebesgue integral of the function f(x) = e^x from 1 to 10 can be calculated similarly to a Riemann integral, as both integrals yield the same result if the function is Riemann integrable on the interval. Additionally, for the Lebesgue integral on the interval X = (0, t), it is true that the derivative of the integral with respect to t equals f(t), except possibly on a set of measure zero, provided that f is summable. This relationship is supported by references to Riesz and Nagy, specifically pages 11 and 47-48.

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eljose
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let be the Lebesgue integral with a meassure [tex]\mu[/tex] then if we call this integral..

[tex]\int_{X}fd\mu=I[/tex]

my questions are..how would you calculate the integral of f(x)=exp(x) from 1 to 10?..another question let be the lebesgue integral on the interval X=(0,t) would be true that:

[tex](\frac{d}{dt}\int_{X}fd\mu=f(t)[/tex]?

thanks...
 
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eljose said:
let be the Lebesgue integral with a meassure [tex]\mu[/tex] then if we call this integral..
[tex]\int_{X}fd\mu=I[/tex]
my questions are..how would you calculate the integral of f(x)=exp(x) from 1 to 10?
The same way you would a Riemann integral. If a function is Riemann integrable on an interval, then it is Lebesque integrable on that interval and the two integrals are the same.

..another question let be the lebesgue integral on the interval X=(0,t) would be true that:
[tex](\frac{d}{dt}\int_{X}fd\mu=f(t)[/tex]?
thanks...
Except possibly on a set of measure 0, yes.
 
you need some hypotheses of course, but if f is "summable" then it is the derivative a.e. of its indefinite integral. see pages 11 and 47, 48 of Riesz Nagy.
 

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