mixted strategy that assigns positive weight to a pure startegy that is dominated
The short answer is that a mixed strategy $\sigma_i$ that uses with positive probability a dominated pure strategy $s_i$ can always be improved by excluding $s_i$ from its mixing support and redistributing its "probability weight" to one of its dominating strategies.

More formally, suppose that strategy $s_i'$ dominates $s_i$ (you can think of $s_i'$ as a pure or mixed strategy itself, this does not make any difference for the argument; for simplicity I suppose $s_i'$ is a pure strategy), that is $\forall s_{-i}\in S_{-i}: \pi(s_i',s_{-i})&gt;\pi(s_i,s_{-i})$.

Also, suppose that strategy $\sigma_i$ plays strategy $s_i$ with positive probability $p_{s_i}\in(0,1)$. Build strategy $\sigma_i'$ as follows: identical to $\sigma_i$ but replace $s_i$ with $s_i'$ , that is play $s_i$ with probability 0 and play $s_i'$ with (additional) probability $p_{s_i}$.

By linearity of the payoff function you have, $\forall s_{-i}\in S_{-i}$:
$$
\pi(\sigma_i',s_{-i})=p_{s_i}\pi(s_i',s_{-i}) + (1-p_{s_i})\pi(\Lambda_i,s_{-i}) &gt; p_{s_i}\pi(s_i,s_{-i}) + (1-p_{s_i})\pi(\Lambda_i,s_{-i}) = \pi(\sigma_i,s_{-i})
$$
where $\Lambda_i$, loosely speaking, denotes the residual combination of strategies played in $\sigma_i$, apart from $s_i$, inherited by $\sigma_i'$ too.