In literature, the biggest class I have seen proven to be in $PP$ is $NP$ (and $coNP$ of course as $PP$ is closed under complementation).However, if we use the following results :$PP^{RP} = PP$  (in fact $PP^{BQP} = PP$)There exists a RP-reduction from $SAT$ to $USAT$Where $USAT$ is the promise class of boolean formulae such thatif $F \in USAT$ then $F$ has a unique solution.if $F \not \in USAT$ then $F$ is not satisfiable.It seems quite straightforward to prove that $\Pi_2^P \subseteq PP$ (hence $\Sigma_2^P \subseteq PP$).Construction:  Let $\mathcal{M}\ $ be the following $PP^{RP}$  Turing Machine :Input: $\forall x_1, \dots, x_m, \exists y_1, \dots y_n, F$Computation: Choose uniformely $x_1, \dots, x_m$.  We obtain then an existential formula.  Use the Valiant-Vazirani theorem to obtain an USAT formula $\exists! z_1, \cdots z_o G$.  We choose uniformly $z_1, \dots, z_o$, and if $G$ is satisfied, accept, otherwise, reject with probability $\frac{1}{2}$.Modify the previous TM by adding $2^n-1$ rejecting paths. (I know I'm going a bit fast, but we can manage that by adjusting rejecting probabilities after having tested G)Validity:if $\forall \dots \exists \dots F$ is true, then $\mathcal{M}\ $ has $2^n$ accepting paths + $2^{n-1}$ rejecting paths + an equal number of accepting and rejecting paths, and then $\mathcal{M}\ $ accepts.if $\forall \dots \exists \dots F$ is false, then $\mathcal{M}\ $ has less than $2^n$ accepting paths + $2^n-1$ rejecting maths + more rejecting paths than accepting paths. then $\mathcal{M}\ $ rejects.Moreover, if we extend Valiant-Vazirani theorem to $RP$-reduce $PH$ to $UPH$ (alternation of forall and unique existentials), then by modifying a bit previous construction we would prove that $PH \subseteq PP$.Is the previous construction correct ?Is there any known litterature around $UPH$ ? (It may not be called $UPH$)
$\mathsf{UPH}$, $\Pi_2^{\mathsf{P}}$ and $\mathsf{PP}$
cc.complexity theory;complexity classes
Your proof must be wrong. All your techniques (including VV) relativize and Beigel has an oracle where $P^{NP}$ is not contained in $PP$. I'm guessing you aren't using VV properly.Also UPH=PH because you can make an existential unique by requiring the lexicographical least witness (using an extra universal quantifier to eliminate the others). 