Integration over the Haar measure of a compact Lie group preserves smoothness? Let $G$ be a compact Lie group. Then there is a unique Haar (probability) measure on $G$. Let $f_g \colon G \to \mathbb{R}$ be a family of smooth functions $(f_g)_{g \in G}$, is the function
$$
G \to \mathbb{R}, \qquad x \mapsto \int_G f_g(x) dg,
$$
smooth?
 A: I guess that the Lie group assumption is unnecessary.
Suppose that $M$ is a smooth manifold, and that $f_y(x)=f(x,y)$ is a smooth function on $M \times M$. You can relax the hypothesis on $f$ just taking a closer look to the proof.
Suppose that a probability measure has been fixed on $M$. Define 
$$F(x) = \int_M f(x,y) dy \ .$$
I claim that $F(x)$ is say of class $C^1$. Here a sketch of the proof.
If $ \gamma : \mathbb{R} \to M$ is a smooth curve ($\gamma(0)=x, \ \dot{\gamma}(0)= \xi$), as $h \to 0$ we have that
$$\frac{F(\gamma(h)) -F(\gamma(0))}{h} = \int_M \frac{f(\gamma(h), y) -f(x, y)}{h} dy \to \int_M \ \frac{d}{dt}\bigg|_{t=0} f(\gamma(t),y) dy = \int_M \big< df_y(x), \xi \big> dy \ ,$$
proving that there exists the derivative
$$ \frac{d}{dt}\bigg|_{t=0} F(\gamma(t)) \ . $$
Here, in order to interchange the limit and the integration procedures you need to use Lebesgue Theorem. This proves that $F$ has continuous partial derivatives in all $x \in M$. On the other hand, if a function has continuous partial derivatives then it is of class $C^1$(this is freshman multivariable calculus).
