Let $(X,\mu,\mathcal{A})$ be a measure space. $f\colon X\to \mathbb{R}$ is $\mu$-measurable if $f$ is a $\mu$-a.e. limit of simple functions $f_j\colon X\to\mathbb{R}$. Functions $f\colon X\to\bar{\mathbb{R}}$, however, are said to be $\mu$-measurable if $f^{-1}(\pm\infty),f^{-1}(O)\in\mathcal{A}$ for all $O$ open in $\mathbb{R}$.
I know the definitions coincide if we consider $f\colon X\to \mathbb{R}$ to be a function with values in $\bar{\mathbb{R}}$. But, if we speak about convergence in $\bar{\mathbb{R}}$, is the pointwise limit $f\colon X\to \bar{\mathbb{R}}$ of a sequence of simple functions $f_j\colon X\to\mathbb{R}$ not necessarily $\mu$-measurable?
The pointwise limit you mention is indeed $\mu$-measurable.
More generally, if $f_n:X\to\bar{\mathbb R}$ is $\mu$-measurable for $n=1,2,\dots$ and a set $A\in\mathcal A$ exists with $\mu(A)=0$ and: $$x\in A^{\complement}\implies \lim_{n\to\infty}f_n(x)\text{ exists in }\bar{\mathbb R}$$ then any function $f:X\to\bar{\mathbb R}$ that equals this limit on $A^{\complement}$ is $\mu$-measurable.
It is not needed here that the $f_n$ are simple and/or take values in $\mathbb R$.