The limit passes under the sign of integral

Question

I have$f:[a,b)\times[c,d]\rightarrow\mathbb{R}, $ (a) $\lim_{y\to y_0} f(x,y)=l(x)$ uniform with respect to $x$ on any compact included in $[a,b) $ and (b) $\int_{a}^{b-0} f(x,y) dx$ is uniform convergent with respect to $y$ on a neighborhood of $y_0$. Then:$\int_a^{b-0}l(x)dx$ is convergent and $\lim_{y\to y_0}\int_a^{b-0}f(x,y)dx= \int_a^{b-0}\lim{y\to y_0}f(x,y)dx $ //I tried to prove it:// Let $\epsilon >0$. from (a) i have: $\exists\delta_\epsilon>0 $ such that $ \forall x\in[a,b) $ and $\forall y \in [c,d] $ with $|y-y_0|<\delta_\epsilon $=>$ |f(x,y)-l(x)|<\epsilon$ now if we put it all under the sign of the integral=>$\int_a^{b-0} |f(x,y)-l(x)|<\int_a^{b-0}\epsilon$ and this show: $\lim_{y\to y_0}\int_a^{b-0}f(x,y)dx= \int_a^{b-0}\lim{y\to y_0}f(x,y)dx $ but i don't used (b) and i feel this is wrong can you help me with a proof please?