= Solution
\b[As literally stated, the claim is false.] Give player one two pure actions, with <payoffs> one and zero against every opposing action. A <mixed strategy> assigning probability one to the first action has exactly that action's <payoffs>, so very weakly dominates it under the printed non-strict inequality. The other pure action nevertheless fails to dominate it.
The intended statement becomes true if the mixture gives positive <probability> to some distinct action, equivalently $p_i<1$. Fix any opposing pure <strategy profile> at which $s_i$ pays one. Since all <payoffs> are in $\{0,1\}$,
$$
1\leq\sum_jp_ju_1(s_j,s_{-1})\leq\sum_jp_j=1.
$$
Equality forces every positive-probability supported action to pay one there. At a profile where $s_i$ pays zero, every other action already pays at least zero. Therefore every action in the <support of a mixed strategy> is no worse than $s_i$ against every opposing profile. Choosing a distinct supported action proves the corrected conclusion: \b[nontrivial mixed <very weak dominance> implies very weak domination by a distinct pure action]. This proof does not insert the missing support condition silently.
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