Lebesgue Integral is Extension of Darboux Integral

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


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

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

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


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

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.