If $h$ is continuous, you don't need any Lebesgue tools.
Since $h$ is continuous then $\forall \epsilon>0$, there exists $\delta>0$ such that if $|y-t|<\delta$, then $|h(x,y)-h(x,t)| < \epsilon$.
Then we have $\frac{1}{2r} \int^{r+t}_{-r+t} h(x,y) dy = h(x,t)+\frac{1}{2r} \int^{r+t}_{-r+t} (h(x,y)-h(x,t)) dy$. If $r<\delta$, we can bound the second term using: $|\int^{r+t}_{-r+t} (h(x,y)-h(x,t)) dy| \leq \int^{r+t}_{-r+t} |(h(x,y)-h(x,t))| dy \leq \epsilon \int^{r+t}_{-r+t} dy = 2 r \epsilon$.
So, if $r < \delta$, we have $|\frac{1}{2r} \int^{r+t}_{-r+t} h(x,y) dy - h(x,t)| \leq \epsilon$. Since $\epsilon>0$ was arbitrary, it follows that the limit is: $\lim_{r\to0^+} \frac{1}{2r} \int^{r+t}_{-r+t} h(x,y) dy = h(x,t).$