Continuity of implicitly defined function Consider a function $g(x)$ defined implicitly via
$\int_{x}^{x + g(x)} f(\xi) d \xi - u(x) = 0$.
I know that for every $x$ a unique g(x) exists.
Furthermore $f$ is locally integrable and $u$ is some continuous function. 
Will $g$ be continuous and how could I prove this?
 A: Yes, $g$ will be continuous.
Defining $G(x)=x+g(x)$ and $F(x)=\int_0^x f(\xi)d\xi$ and $U(x)=F(x)+u(x)$, your equation becomes
$$F(G(x))=U(x),$$
with $F$ and $U$ continuous. The hypothesis is that there is a unique $G$ satisfying this equation, and the question is whether $G$ must be continuous.
Consider the restriction $\hat F:G(\mathbb R)\to U(\mathbb R)$. This is either strictly increasing or strictly decreasing, and $G(\mathbb R)$ is connected. To see this, consider $y_1<y_2<y_3$ with $F(y_1)<F(y_3)$ and $y_1,y_3\in G(\mathbb R)$. If $F(y_2)<F(y_1)$ then by the intermediate value theorem $F(y_1)\in F([y_2,y_3])$, contradicting the uniqueness of $G$. Similarly $F(y_2)>F(y_3)$ is impossible, so $F(y_1)<F(y_2)<F(y_3)$. This applies to all $y_2\in [y_1,y_3]$, so $F(y_2)\in [F(y_1),F(y_3)]\subseteq U(\mathbb R)$, so $y_2\in G(\mathbb R)$ by uniqueness of $G$. A similar argument applies when $F(y_1)>F(y_3)$.
It follows that $\hat F$ is an open map: the image of a relatively open interval in $G(\mathbb R)$ is relatively open in $U(\mathbb R)$. So $G=\hat F^{-1}\circ U$ is continuous.
