# 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.

$\blacksquare$

## Sources

- 2005: René L. Schilling:
*Measures, Integrals and Martingales*... (previous) ... (next): $11.8 \ \text{(i)}$