Lebesgue Integral is Extension of Riemann Integral

Theorem

If $f: \left[{a . . b}\right] \to \R$ is Riemann integrable, then it is measurable, and:

$\displaystyle R \int_a^b f \left({x}\right) \mathrm d x = \int_a^b f$

where $\displaystyle R \int_a^b$ is the Riemann integral and the $\displaystyle \int_a^b$ is the Lebesgue integral.

Proof

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

$\displaystyle L \left({P}\right) \le \sup_{\phi \le f} \int_a^b \phi \left({x}\right) \mathrm d x \le \inf_{\psi \ge f} \int_a^b \psi \left({x}\right) \mathrm d x \le U \left({P}\right)$

where $L \left({P}\right)$ and $U \left({P}\right)$ are the lower sum and upper sum as defined in the definition of definite integral.

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

$\blacksquare$