To start with, let us consider the following
Definition
Let $f:\Omega\to\overline{\mathbb{R}}_{\geq0}$ be a nonnegative measurable function on $(\Omega,\mathcal{F},\mu)$. The integral of $f$ with respect to $\mu$, also denoted by $\int fd\mu$, is defined as
\begin{align*}
\int fd\mu = \lim_{n\to\infty}\int f_{n}d\mu
\end{align*}
where $\{f_{n}\}_{n\geq 1}$ is any sequence of nonnegative simple functions such that $f_{n}(\omega)\uparrow f(\omega)$ for all $\omega$.
Definition
Let $f$ be a real valued measurable function on a measure space $(\Omega,\mathcal{F},\mu)$. Let $f^{+} = fI_{\{f\geq0\}}$ and $f^{-} = – fI_{\{f<0\}}$. The integral of $f$ with respect to $\mu$, denoted by$\int fd\mu$, is defined as
\begin{align*}
\int fd\mu = \int f^{+}d\mu – \int f^{-}d\mu
\end{align*}
provided that at least one of the integrals on the right side is finite.
Thus the problem to integrate a measurable function is reduced to calculate the integral of nonnegative measurable functions.
Moreover, we also have the following
Theorem
Let $f$ be a bounded function on a bounded interval $[a,b]$. Then $f$ is Riemann integrable on $[a,b]$ iff $f$ is continuous a.e. according to the Lebesgue measure $m$ on $[a,b]$. In this case, $f$ is Lebesgue integrable on $[a,b]$ and the Lebesgue integral $\int_{[a,b]}fdm$ equals the Riemann integral $\int_{[a,b]}f$, i.e., the two integrals coincide.
My question
In the case of Riemann integrable functions, one may apply the Fundamental Theorem of Calculus to calculate integrals, since antiderivatives and integration are strongly related.
My question is: how do we calculate the integral of measurable functions which are not Riemann integrable? Is there a routine method to do so other than the direct application of its definition?
I am new to measure theory, so any contribution is appreciated.
Best Answer
In the book Brian S. Thomson, Judith B. Bruckner, Andrew M. Bruckner "REAL ANALYSIS" Second Edition (2008) on page 360 there is following theorem:
If $f$ have bounded derivative on $[a,b]$, then for Lebesgue integral holds $$f(b)-f(a) = \int\limits_{a}^{b}f^{'}d\lambda$$