Let $\mu$ be a real Radon measure over $\mathbb{R}$. Its total variation norm is given by \begin{equation} \lVert \mu \rVert_{\mathcal{M}(\mathbb{R}, \mathbb{R})} = \sup_{f \in \mathcal{C}_0(\mathbb{R}, \mathbb{R}), \lVert f \rVert_\infty \leq 1} | \langle \mu , f \rangle | \end{equation} where $\mathcal{C}_0(\mathbb{R}, \mathbb{R})$ is the space of real valued continuous functions vanishing at infinity.
Looking at $\mu$ as a complex Radon measure, its total variation norm is now \begin{equation} \lVert \mu \rVert_{\mathcal{M}(\mathbb{R}, \mathbb{C})} = \sup_{f \in \mathcal{C}_0(\mathbb{R}, \mathbb{C}), \lVert f \rVert_\infty \leq 1} | \langle \mu , f \rangle | \end{equation} where $\mathcal{C}_0(\mathbb{R}, \mathbb{C})$ is the space of complex valued continuous functions vanishing at infinity.
Clearly, since $\mathcal{C}_0(\mathbb{R}, \mathbb{R}) \subset \mathcal{C}_0(\mathbb{R}, \mathbb{C})$, one has that $\lVert \mu \rVert_{\mathcal{M}(\mathbb{R}, \mathbb{R})} \leq \lVert \mu \rVert_{\mathcal{M}(\mathbb{R}, \mathbb{C})}$.
Question: Do we have that the real and complex norms are equal, i.e., $\lVert \mu \rVert_{\mathcal{M}(\mathbb{R}, \mathbb{R})} = \lVert \mu \rVert_{\mathcal{M}(\mathbb{R}, \mathbb{C})}$ for any real Radon measure?
This is a standard trick in complex analysis; pick a complex function for which $| \langle \mu , f \rangle |=a>0$; this means that there is $b \in [0, 2\pi]$ st $ \langle \mu , e^{ib}f \rangle =a$; let $g=e^{ib}f$ and conjugating and using that $\mu$ real, we get:
$\langle \mu , g \rangle =a$ and $\langle \mu , \bar g \rangle =a$ so summing:
$\langle \mu , \Re g \rangle =a$, hence we are done by letting $a_n \to \lVert \mu \rVert_{\mathcal{M}(\mathbb{R}, \mathbb{C})}, f_n$ apriori complex realizing the $a_n$ being replaced with the real functions $\Re g_n$ constructed above (and noting that all the required properties of $f$ are preserved by $g, \Re g$)