Let $f:\mathbb{R}\longrightarrow \mathbb{R}$ be a Borel measurable function. Then $$ \lim_{\lambda\to\infty}\int_{\mathbb{R}}f(x)e^{i\lambda x}d\mu(x)=0. $$ I obtain this result by showing that it is true for $f=\sum_{k=1}^nb_k\chi_{(a_k,b_k)}$. And the collection of simple functions of the form $f=\sum_{k=1}^nb_k\chi_{(a_k,b_k)}$ is dense in Borel mesurable functions on $\mathbb{R}$.
Is there any counter-example that $f$ is only Lebesgue measurable but not Borel, and $$ \lim_{\lambda\to\infty}\int_{\mathbb{R}}f(x)e^{i\lambda x}d\mu(x)\neq 0? $$
What you should be using is that the set of simple functions you consider is dense in the space $L^1(\Bbb{R}^n)$.
You will also have to use that the Fourier transform $\mathcal{F} : L^1 \rightarrow C_b$ is bounded (where $C_b$ is the space of bounded continuous functions with the sup-norm). You will also want to use that $C_0 \subset C_b$ is closed, where $C_0$ is the space of continuous functions vanishing at infinity.
The space of all Borel measurable functions has no natural norm in which your statement it true.
Main point Finally note that for every Lebsgue measurable function $f$, there is a Borel measurable function $g$ with $f=g$ a.e., so that $\hat{f} = \hat{g}$.
Thus it makes no difference which notion of measurability you take.