I wish to show that in case $f:\mathbb{R^2}\rightarrow\mathbb{R}^2$ is continuous then for $g(t) :[a,b]\rightarrow \mathbb{R^2}$ , defined as: $$g(t) = f(z+ht),\quad z,h\in \mathbb{R}^2,$$ then $$\lim_{h\rightarrow 0} \int_{t=a} ^{t=b} f(z +h\cdot t)dt = \int_{t=a} ^{t=b} \lim_{h\rightarrow 0} f(z +h\cdot t)dt.$$
But I am not sure what is the definition of the Riemann integral for function $g:[a,b]\subseteq\mathbb{R} \rightarrow \mathbb{R^2}$.
I wish for a reference which explains Riemann's integration for this and similar cases.
Let $z,h$ be fixed. By the mean value theorem for integrals, there is $t_h \in [a,b]$ such that
$\int_{t=a} ^{t=b} f(z +h\cdot t)dt=(b-a)f(z+ht_h)$. Since $a \le t_h \le b$ , we have $ht_h \to 0$ as $h \to 0$, thus $f(z+ht_h) \to f(z)$ as $h \to 0$.
Consequence: $\lim_{h\rightarrow 0} \int_{t=a} ^{t=b} f(z +h\cdot t)dt=(b-a)f(z)$.
On the other hand: $\lim_{h\rightarrow 0} f(z +h\cdot t)=f(z)$ for all $t \in [a,b]$, thus
$ \int_{t=a} ^{t=b} \lim_{h\rightarrow 0} f(z +h\cdot t)dt= \int_{t=a} ^{t=b} f(z) dt=(b-a)f(z).$