Given $f_k=\big(\frac{x}{k}+1\big)^{-k}\sin(\frac{x}{k})$, assuming x>0.
It seems the question is trivial, since $\big(\frac{x}{k}+1\big)^{-k}$ and $\sin(\frac{x}{k})$ are continuous for fixed $x$, they are measurable. And $f_k\to 0$ since $\sin(\frac{x}{k})\to 0$.
To find a bound, we have $\big(\frac{x}{k}+1\big)^{-k}\sin(\frac{x}{k})\leq \big(\frac{x}{k}+1\big)^{-k}\leq 1$. So we can take the constant function $g=1$, and it is measurable by the same logic.
is this correct? or I am completely off? thanks for the help