Triangle inequality for integrals of complex functions of real variable If $f:[a, b] \to \mathbb{C}$ is an integrable function on a real interval, then
$$\left| \int_a^b f \right | \le \int_a^b |f| $$
There's a proof of this in Ahlfors' Complex Analysis book which really cool: you just rotate things so that the integral is real by multiplying by $e^{i\theta}$, then apply the similar result for real-valued functions.
Qeustion: Can this also be done by something like applying the discrete triangle equality to Riemann sums for partitions and passing through the limit? If so, is there a good reference for it?
Thank you?
 A: Riemann-integrable functions have the nice property that for every sequence of partitions $P_n = \{ a = a_0^{(n)} < a_1^{(n)} < \dotsc < a_{r_n}^{(n)} = b\}$ of the interval $[a,b]$ whose mesh $\mu(P_n) = \max \{ a_\rho^{(n)} - a_{\rho-1}^{(n)} : 1 \leqslant \rho \leqslant r_n\}$ tends to zero, and every choice of support points $\xi_\rho^{(n)} \in \left(a_{\rho-1}^{(n)},\, a_\rho^{(n)}\right)$, the sequence of Riemann sums
$$R\left(P_n,(\xi^{(n)}),f\right) := \sum_{\rho = 1}^{r_n} f\left(\xi_\rho^{(n)}\right)\cdot (a_\rho^{(n)} - a_{\rho-1}^{(n)})$$
converges to the integral $\int_a^b f(x)\,dx$. That holds for real-valued, complex-valued, or $\mathbb{R}^n$-valued functions. Also, for properly Riemann-integrable functions (defined, bounded, and Riemann-integrable on a compact interval), the absolute value $\lvert f\rvert$ is Riemann-integrable too.
Thus, fixing any sequence of partitions and associated sequence of support points, applying the triangle inequality to each Riemann sum for our chosen sequence, we have
$$\begin{align}
\left\lvert R\left(P_n,(\xi^{(n)}),f \right)\right\rvert
&= \left\lvert \sum_{\rho = 1}^{r_n} f\left(\xi_\rho^{(n)}\right)\cdot (a_\rho^{(n)} - a_{\rho-1}^{(n)})\right\rvert\\
&\leqslant \sum_{\rho=1}^{r_n} \left\lvert f(\xi_\rho^{(n)})\right\rvert \left( a_\rho^{(n)} - a_{\rho-1}^{(n)}\right)\\
&= R\left(P_n,(\xi^{(n)}),\lvert f\rvert \right),
\end{align}$$
and taking the limit we obtain
$$\left\lvert \int_a^b f(x)\,dx \right\rvert \leqslant \int_a^b \left\lvert f(x)\right\rvert\,dx.$$
We didn't use any particular properties of $\mathbb{C}$ here, so the same argument, replacing the absolute modulus with the norm, yields
$$\left\lVert \int_a^b f(x)\,dx \right\rVert \leqslant \int_a^b \left\lVert f(x)\right\rVert\,dx$$
for any Riemann-integrable function with values in some $\mathbb{R}^n$, endowed with whichever (semi-)norm.
A: The Riemann sums use the supremum and the infimum of $f$ over subintervals of the partition. But the $\sup f$ and $\inf f$ do not make sense if $f$ is complex valued, since $\mathbb{C}$ is not ordered.
