We say that a sequence $(\mu\_{n})$ of measures in $M\_{b}(Q)$ converges **tightly** (or, equivalently, in the narrow topology of measures) to a measure $\mu$ in $M\_{b}(Q)$ if
$$\lim\_{n\to\infty}\int\_{Q}\varphi d\mu\_{n} \to \int\_{Q}\varphi d\mu\quad (1.1)$$
for every $\varphi\in C\_{b}(Q)$. If (1.1) holds only for all the continuous functions $\varphi$ with compact support in $Q$ (i.e., $\varphi\in C\_{c}(Q))$, then we have the usual weak–\* convergence in $M\_{b}(Q)$. What is the relation between **tight** and strong convergence? The two definitions are equivalent for nonnegative measures, but why not for general signed measures (a counter-example))?



Convergence in the narrow topology of measures and strongly converge for signed measures
You claim that weak (i.e. tight) and strong convergence are equivalent for non-negative measures, but this is false: take $Q = (0,T) \times (-1,1)$, let $t \in (0,T)$ and let $\mu\_n$ be the Dirac measure concentrated at $(t, \frac 1 n)$ and $\mu$ be the Dirac measure concentrated at $(t,0)$. If $\varphi \in C\_b (Q)$ then, since $\varphi$ is continuous,


$$\int \_Q \varphi \ \mathrm d \mu\_n = \varphi \left( t, \frac 1 n \right) \to \varphi (t,0) = \int \_Q \varphi \ \mathrm d \mu$$


so that $\mu\_n \to \mu$ tightly.


On the other hand, $\{(t,0)\}$ is measurable, and


$$\mu\_n \big( \{(t, 0)\} \big) = 0 \not\to 1 = \mu \big( \{(t, 0)\} \big)$$


which shows that $\mu\_n \not\to \mu$ strongly.


