To summarize the situation. Let $(X,\mathcal{F})$ a measurable space.
The space $M(X,\mathcal{F})$ of all real-valued signed measures on $(X,\mathcal{F})$ is a Banach space wrto the total variation norm $\|\mu\|:=|\mu|(X)$ where  the (non-negative) measure $|\mu|:= \mu_+ +\mu_-$ is the variation of $\mu$. The space $M(X,\mathcal{F})$may be isometrically embedded as a norm-closed subspace of the dual space $B(X,\mathcal{F})^*$ of the Banach space $B(X,\mathcal{F})$ of all bounded measurable functions on $X$,  $B(X,\mathcal{F})$ with the uniform norm $\|\cdot\|_\infty$ ( that dual is in general much larger, since it also contains all additive measures). Note that the space of simple functions $S(X,\mathcal{F})$, linear span of characteristic functions of measurable sets, is norm dense in $B(X,\mathcal{F})$ via the usual approximation $f_n(x):= \lfloor nf(x)\rfloor/n$.

So the above "strong convergence of measures", that is with test functions in $B(X,\mathcal{F})$, is the weak* convergence of   $B(X,\mathcal{F})^*$, in the particular cas eof sequences in  $M(X,\mathcal{F})$. In particular any such convergent sequence of measures is norm bounded (as any w* convergent sequence of elements in a dual space), and it is "weakly convergent" in the above sense, that is with characteristic functions, hence also with simple functions as test.  It is not norm convergent in general (as an example, take as said e.g. $\mu_n$ absolutely continuous w.r.to the Lebesgue measure on $X:=[0,1]$
and with densities $g_n\in L^1$ weakly convergent but not norm convergent). 

Conversely, a norm bounded sequence of measures $\mu_n$ that weakly converges to $\mu$ wrto test functions $g$ in $S(X,\mathcal{F})$ also converges with test functions $f$ in $B(X,\mathcal{F})$, again an elementary and general fact. You can see it writing
$$\langle \mu_n, f\rangle - \langle\mu, f\rangle=\langle\mu_n-\mu, g\rangle+ \langle\mu_n-\mu, g -f\rangle $$
with a simple function $g$, so that
$$\limsup_{n\to\infty}|\langle\mu_n, f\rangle-\langle\mu, f\rangle|\le \limsup_{n\to\infty}|\langle\mu_n-\mu, g\rangle|+ \big(\sup_n \|\mu_n\|+\|\mu\|\big)\|g -f\|_\infty $$
whence
$$\limsup_{n\to\infty}|\langle\mu_n, f\rangle-\langle\mu, f\rangle|=0,$$

since simple functions are  $\|\cdot\|_\infty$ dense.