📂Measure Theory

Lebesgue Integral as a Generalization of the Riemann Integral

Theorem ¹

Let $f : [a,b] \to \mathbb{R}$ and $g : \mathbb{R} \to [0,\infty)$ be bounded functions.

• [1]: $f$ being Riemann integrable on $[a,b]$ is equivalent to $f$ being continuous almost everywhere on $[a,b]$ with respect to the Lebesgue measure.
• [2]: If $\displaystyle \int_{a}^{b} f(x) dx$ exists, then $\displaystyle \int_{a}^{b} f(x) dx = \int_{[a,b]} f dm$
• [3]: If $\displaystyle \int_{-\infty}^{\infty} g(x) dx$ exists, then $\displaystyle \int_{-\infty}^{\infty} g(x) dx = \int_{\mathbb{R}} g dm$

Explanation

All those numerous discussions about measure may well be regarded as being for the sake of this ‘generalization of the integral’. It is a good thing that the Lebesgue integral lets us compute the definite integrals of more functions, but it would be meaningless if its value differed from the Riemann integral.

With elementary analysis it was difficult to determine whether a function is Riemann integrable, but with Theorem [1] the proof becomes very easy. For example, the Dirichlet function $\mathbb{1}_{\mathbb{Q}}$ is discontinuous at every point of $[0,1]$, so its Riemann integral does not exist.