Lebesgue Integral is Extension of Riemann Integral

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Theorem

Let $f: \left[{a \,.\,.\, b}\right] \to \R$ be a Riemann integrable function.

Then it is also Lebesgue integrable, and furthermore:

$\displaystyle R \int_a^b f \left({x}\right) \, \mathrm d x = \int_{\left[{a \,.\,.\, b}\right]} f \, \mathrm d \lambda$

where $\displaystyle R \int_a^b$ is the Riemann integral and $\displaystyle \int_{\left[{a \,.\,.\, b}\right]}$ 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.


$\blacksquare$


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