A new result of implicit function (Theorem M-R) Let $f\in C^0(\mathbb R^2, \mathbb R)$, so that there exists an $M>0 \in \mathbb R$ with $f(x,y) \geq M$ for all $(x,y)\in\mathbb R^2$.
Does it follow for any $g \in C^0(\mathbb R,\mathbb R)$ that there exsists a unique $ h \in C^0(\mathbb R,\mathbb R)$ so that
$$\int_0^{h(a)} f(a,x)\,\text{d}x=g(a)$$
for all $a\in\mathbb R$?
 A: Existence and unicity have been prooved in comments, let's prove continuity.
Take $a\in \mathbf R$ and assume $h(a) \neq 0$. The function $f$ is continuous on $\mathbf R^2$ and thus uniformly continuous on the compact set $K=[a-1;a+1]\times [0;h(a)]$. This means that for every $\varepsilon>0$ there exists some $1>\delta>0$ such that $|a-b|<\delta \Rightarrow |f(a,x)-f(b,x)|<\varepsilon/|h(a)|$ for every $x$ in $[0;h(a)]$ and thus
$$|a-b|<\delta \Rightarrow \left|\int_0^{h(a)}f(a,x)dx-\int_0^{h(a)}f(b,x)dx\right|<\varepsilon.$$ 
Now take $\delta$ to be small enough so that $|g(a)-g(b)|<\varepsilon$ and notice that 
$$|g(a)-g(b)|=\left|\int_0^{h(a)}f(a,x)dx-\int_0^{h(b)}f(b,x)dx\right|=\left|\int_0^{h(a)}f(a,x)dx-\int_0^{h(a)}f(b,x)dx -\int_{h(a)}^{h(b)}f(b,x)dx\right|.$$
Since $f\geq M>0$ everywhere we get that 
$$|h(a)-h(b)|> 2\varepsilon/M \Rightarrow \left|\int_{h(a)}^{h(b)}f(b,x)dx \right|>2\varepsilon$$ 
which in turn would imply $|g(a)-g(b)|>2\varepsilon-\varepsilon=\varepsilon$, contradicting the assumption made earlier. So the only possibility is that $|h(a)-h(b)|<2\varepsilon/M$ and thus $h$ is continuous. The case $h(a)=0$ is treated in the exacte same way except that we don't have to bound the term $\left|\int_0^{h(a)}f(a,x)dx-\int_0^{h(a)}f(b,x)dx\right|$ since it's equal to $0$.
A: As Renart, I'll ignore existence and uniqueness, and focus on continuity. However, I'll use a different (but equivalent) definition of continuity : a function is continuous if and only if the pre-image of any open ball is open. That avoids $\varepsilon-\delta$ arguments, although this proof can be adapted to use the $\varepsilon-\delta$ definition if needed.

First point : for all $M \in \mathbb{R}$, the function $I_M : a \mapsto \int_0^M g(a,x) dx$ is continuous.
Proof(s) : I'll give two proofs.
1) Let $a$, $M \in \mathbb{R}$. The subspace $[a-1, a+1]\times [0,M]$ is compact and $f$ is continuous, so $f$ is uniformly continuous on $[a-1, a+1]\times [0,M]$. Hence, there exists a modulus of continuity $\omega$ for $f$ on this subspace ; that is, for all $x$, $y \in [a-1, a+1]\times [0,M]$,
$$|f(x)-f(y)| \leq \omega (\|x-y\|).$$
Then, we get, for all $a' \in [a-1,a+1]$:
$$\left| \int_0^M f(a,x) dx - \int_0^M f(a',x) \right| \leq \int_0^M |f(a,x)-f(a',x)| dx \leq M \omega (|a-a'|).$$
In particular, the differences converges to $0$ when $a'$ converges to $a$.
2) Since $[a-1, a+1]\times [0,M]$ is compact and $f$ is continuous, $f$ is bounded on $[a-1, a+1]\times [0,M]$. Let $(a_n)_n$ be a sequence which converges to $a$. Then:
$$\lim_{n \to + \infty} \int_0^M f(a_n,x)dx = \int_0^M \lim_{n \to + \infty} f(a_n,x)dx =\int_0^M f(a,x)dx,$$
where we used first the dominated convergence theorem, and then the continuity of $f$.

Now, I'll prove the claim. Let $y \in \mathbb{R}$ and $\varepsilon >0$. Then:
$$h^{-1} (B(y,\varepsilon)) = \left\{a : y-\varepsilon < h(a) \right\} \cap \left\{a : h(a) < y+\varepsilon \right\}$$
But, since $M \mapsto \int_0^M f(a,x) dx$ is strictly increasing, 
we have $\left\{a : y-\varepsilon < h(a) \right\} = \left\{a : \int_0^{y-\varepsilon} f(a,x) dx -g(a) < 0 \right\} = (I_{y-\varepsilon}-g)^{-1} (\mathbb{R}_-^*)$.
Since $I_{y-\varepsilon}-g$ is continuous by the fist point, $\left\{a : y-\varepsilon < h(a) \right\}$ is open. By the same argument, $\left\{a : y+\varepsilon > h(a) \right\}$ is open. Hence, as the intersection of two open sets is open, $h^{-1} (B(y,\varepsilon))$ is open. Since this is true for all $y \in \mathbb{R}$ and all $\varepsilon >0$, it follows that $h$ is continuous.
