Lebesgue Integral is Extension of Darboux Integral

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Theorem

Let $f: \closedint a b \to \R$ be a Darboux integrable function.

Then it is also Lebesgue integrable, and furthermore:

$\ds R \int_a^b \map f x \rd x = \int_{\closedint a b} f \rd \lambda$

where $\ds R \int_a^b$ is the Darboux integral and $\ds \int_{\closedint a b}$ is the Lebesgue integral.


Proof

Since every step function is also a simple function, we have

$\ds \map L P \le \sup_{\phi \mathop \le f} \int_a^b \map \phi x \rd x \le \inf_{\psi \mathop \ge f} \int_a^b \map \psi x \rd x \le \map U P$

where $\map L P$ and $\map U P$ are the lower sum and upper sum as defined in the definition of definite integral.


Since $f$ is Darboux integrable, the inequalities are all equalities and $f$ is measurable by basic properties of measurable functions.


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$\blacksquare$


Sources

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