Approximating common knowledge with common beliefs (Monderer and Samet, 1989)
Showing the left-hand inequality

Let's rearrange the equation,
$$
\begin{align*}
x &amp;= r_i \frac{\mu\left( B^p_i\left( E \right) \right)}{\mu\left( E \right)} - \frac{\mu\left( X \cap \left[ B^p_i\left( E \right) \setminus E \right] \right)}{\mu\left( E \right)} \\
&amp; \iff \underbrace{x \mu\left( E \right) + \mu\left( X \cap \left[ B^p_i\left( E \right) \setminus E \right] \right)}_{\text{LHS}} = r_i \mu\left( B^p_i\left( E \right) \right).
\end{align*}
$$
Since all terms in $\text{LHS}$ are weakly positive, we can omit $\mu( X \cap [ B^p_i( E ) \setminus E ] )$ from $\text{LHS}$, giving
$$
x \mu\left( E \right) \leq r_i \mu\left( B^p_i\left( E \right) \right).
$$
Since $\mu( E ) \geq p \mu( B^p_i( E ) )$, this directly implies $x p \leq r_i$.

Showing the right-hand inequality

Since $\mu( E ) \leq \mu( B^p_i( E ) )$ and $x \geq 0$, we have
$$
\begin{align*}
&amp;\frac{1}{\mu\left( E \right)} \left[ r_i \mu\left( B^p_i\left( E \right) \right) - \mu\left( X \cap \left[ B^p_i\left( E \right) \setminus E \right] \right) \right] \\
&amp;\geq \frac{1}{\mu\left( B^p_i\left( E \right) \right)}\left[ r_i \mu\left( B^p_i\left( E \right) \right) - \mu\left( X \cap \left[ B^p_i\left( E \right) \setminus E \right] \right) \right] \\
&amp;= \underbrace{r_i - \frac{\mu\left( X \cap \left[ B^p_i\left( E \right) \setminus E \right] \right)}{\mu\left( B^p_i\left( E \right) \right)}}_{\text{RHS}}.
\end{align*}
$$
It must be that $X \cap [ B^p_i( E ) \setminus E ] \subseteq B^p_i( E ) \setminus E$, so
$$
\text{RHS} \geq r_i - \frac{\mu\left( B^p_i\left( E \right) \setminus E \right)}{\mu\left( B^p_i\left( E \right) \right)} = r_i - \frac{\mu\left( B^p_i\left( E \right) \right) - \mu\left( E \right)}{\mu\left( B^p_i\left( E \right) \right)} = r_i - \left( 1 - \mu\left( \left. E \right| B^p_i\left( E \right) \right) \right).
$$
By definition (and this is inequality (9) in the text) $\mu( E | B^p_i( E ) ) \geq p$, thus
$$
\text{RHS} \geq r_i - \left( 1 - p \right).
$$
Noting that the original equality was $x = \text{RHS}$, we have
$$
x \geq r_i - \left( 1 - p \right) \;\; \iff \;\; r_i \leq x + 1 - p.
$$