Let $g:I \rightarrow \mathbb{R}$ be a bounded function so there exists a sequence of step functions $g_{i}: I \rightarrow \mathbb{R}, i \in \mathbb{N}$ with $$\lim_{i \to \infty}\Big(\sup_{x \in I}|g_{i}(x)-g(x)|\Big)=0$$.
Show that $g$ is $B_{I}$-$B(\mathbb{R})$ measurable where $B_{I}$ denotes the Borel sigma algebra restricted to $I, B(\mathbb{R})$ denotes the Borel sigma algebra and that $\lim_{i \to \infty}\int_{I}g_{i}d\mu = \int_{I}gd\mu$.
Attempt:
To show it is measurable I need to show $\forall A \in B(\mathbb{R}), g^{-1}(A) \in B_{I}$, or alternatively one of the other equivalent forms of measurability. Since $g$ is bounded I am thinking that perhaps the form $g^{-1}([-\infty,a]) \in B_{I}, \forall a \in \mathbb{R}$ could be a good approach but I am unsure how to proceed.
For the integral is it an application of Dominated Convergence Theorem? so I can just bring the limit inside the integral and then the result follows...