- #1

AxiomOfChoice

- 533

- 1

[tex]

F(x) = \frac{d}{dx} \int_a^x F'(y)dy

[/tex]

to hold, where "[itex]\int[/itex]" is the Lebesgue integral?

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- Thread starter AxiomOfChoice
- Start date

- #1

AxiomOfChoice

- 533

- 1

[tex]

F(x) = \frac{d}{dx} \int_a^x F'(y)dy

[/tex]

to hold, where "[itex]\int[/itex]" is the Lebesgue integral?

- #2

mathman

Science Advisor

- 8,069

- 543

Your question is confusing (also misstated). You have F = F'.

- #3

AxiomOfChoice

- 533

- 1

Your question is confusing (also misstated). You have F = F'.

You're right. Sorry. What I meant to ask was what are necessary and sufficient conditions for the equality

[tex]

F(x) = \frac{d}{dx} \int_a^x F(y)dy

[/tex]

to hold, when the integral is the Lebesgue (not the Riemann) integral.

- #4

snipez90

- 1,101

- 5

Anyways, the first relevant theorem is the following:

(*)The indefinite integral [itex]F(x) = \int_{a}^{x}f(t)\,dt[/itex] of a Lebesgue integrable function f is absolutely continuous.

This direction is pretty easy to prove, and the relevant idea is the continuity of the Lebesgue integral.

The other relevant theorem, which is apparently due to Lebesgue, is the following:

(**)If F is absolutely continuous on [a,b], then the derivative F' is Lebesgue integrable on [a,b], and

[tex]F(x) = F(a) + \int_{a}^{x}F'(t)\,dt.[/tex]

At least the way I learned it, the bulk of this proof rests in the following lemma:

If f is an absolutely continuous nondecreasing function on [a,b] such that f'(x) = 0 almost everywhere, then f is constant.

Combining (*) and (**) gives you a possible characterization of what you're looking for. If you're unclear about the relevant ideas involved, wikipedia is probably your best bet. Feel free to ask me about any particular definitions that might need clarification, since I know Kolmogorov & Fomin sometimes use definitions that are different but equivalent to those found in more modern texts.

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