Continuity of an integral with respect to one variable Let $V\subseteq \mathbb{R}^n$ and $f:V\to\mathbb{R}^n$. Consider the function
$$g(x_1,x_2,...,x_n) = \int_{x_2}^{x_1} {f(t,x_2,...,x_n)dt}$$
on $V$. What conditions will I need to conclude that $g$ is also continuous on $V$? (I would also appreciate if you give me some justification for them.)
In the problem I am working on, I know for certain that $f$ is continuous and Lipschitz on $V$. But all I can deduce from this is that if I fix $x_2,...,x_n$, then $g$ (as a function of $x_1$ alone) is continuous. Will it help that $V$ is also compact, so that $f$ is uniformly continuous there?
If you are interested, I encounter this problem in the proof of a theorem in Coddington and Levinson's "Theory of differential equations." In particular, it is in the proof of Theorem 7.1 on pages 23-24. The successive approximations there are defined similarly as above; he says that they are continuous without much detail. 
 A: I am assuming you are using the Euclidiean norm $|.|$ on $V$. Let $x=(x_1,x_2,...,x_n)\in V$, and $\epsilon>0$. Since $f$ is continuous, there exists $\sigma>0$ such that 
$$
    |x-y|\leq\sigma \implies |f(x)-f(y)|\leq\epsilon
$$
Now consider $y\in V$ such that $|x-y|\leq\sigma$. For any $t$ between $x_1$ and $x_2$, we have $|(t,x_2,...,x_n)-(t,y_2,...,y_n)|\leq|x-y|\leq\sigma$, and therefore $|f(t,x_2,...,x_n)-f(t,y_2,...,y_n)|\leq\epsilon$
\begin{align}
 |g(x)-g(y)|&=\left|\int_{x_2}^{x_1}f(t,x_2,...,x_n)\mathrm{d}t
   -\int_{y_2}^{y_1}f(t,y_2,...,y_n)\mathrm{d}t \right|\\
  &=\left|\int_{x_2}^{x_1}(f(t,x_2,...,x_n)-f(t,y_2,...,y_n))\mathrm{d}t\right.\\
  &\kern 3em \left.+\int_{x_2}^{y_2}f(t,y_2,...,y_n)\mathrm{d}t
   +\int_{y_1}^{x_1}f(t,y_2,...,y_n)\mathrm{d}t
   \right|\\
  &\leq|x_1-x_2|\epsilon+\left|\int_{x_2}^{y_2}f(t,y_2,...,y_n)\mathrm{d}t\right|+\left|\int_{y_1}^{x_1}f(t,y_2,...,y_n)\mathrm{d}t\right|
\end{align}
Since $f$ is continuous on the compact $V$, it is also bounded, so there exists $M>0$ such that for all $x\in V$, $|f(x)|\leq M$. Finally, we find:
$$
    |g(x)-g(y)|\leq|x_1-x_2|\epsilon+(|y_2-x_2|+|y_1-x_1|)M.
$$
Choosing $y$ such that $|y-x|\leq\min(\sigma,\epsilon)$ yields
$$
    |g(x)-g(y)|\leq(|x_1-x_2|+2M)\epsilon,
$$
proving the continuity at point $x$. This reasoning is valid for any $x$ in $V$, and $g$ is therefore continuous on $V$.
