Let $g: \mathbb{C}\times[a,b]\to\mathbb{R}$ a continuous function and
$$g(t)=g(h_0,t)=\lim\nolimits_{h\to h_0} g(h,t) \ \forall t\in \mathbb{R}$$
Is the following result/reasoning correct ? If we have a sequence of functions, we need uniform convergence to pass the limit inside the integral, but it seems it works everytime here. Or there is a flaw ?
We have
$$\lim_{h\to h_0} \int_a^b g(h,t)dt= \int_a^b g(t)dt$$
because
Idea 1:
$$\forall t\in[a,b], \ \forall > \epsilon >0, \exists \delta(t)>0\text{ > such that }|g(h,t)-g(t)|<\epsilon \ > \forall |h-h_0|<\delta(t)$$
Then if $|h-h_0|<\delta=\inf_{t\in[a,b]} > \delta(t)$,
$$|g(h,t)-g(t)|<\epsilon \ \forall > |h-h_0|<\delta(t), \ \forall > t\in[a,b]$$
Thus
$$\left|\int_a^b > (g(h,t)-g(t))dt\right|\le \int_a^b > |g(h,t)-g(t)|dt\le (b-a)\epsilon$$
Doesn't work if $\inf_{t\in[a,b]} > \delta(t)=0$, see George's comment.
~
Idea 2: (added after George's comment)
Let $\overline{B(h_0, \gamma)}\subset > C$ a closed ball. Resctrict $g$ to $ > \overline{B(h_0, \gamma)}\times[a,b]$. Because it is continuous on a compact, this function will be uniformly continuous. Thus
$$\forall \epsilon >0 \ \exists > \delta>0 \ : \ > |g(h,t)-g(h_0,t)|<\epsilon \ \forall > |h-h_0|<\min(\delta, \gamma)$$
and
$$\left|\int_a^b > (g(h,t)-g(t))dt\right|\le \int_a^b > |g(h,t)-g(t)|dt\le (b-a)\epsilon$$
if $|h-h_0|<\min(\delta, \gamma)$.
Does this one work ;) ?
Thank you for any help :)