Since $a \leq f \leq b$ almost everywhere, we may apply results
to conclude
\begin{equation}
\int_E f \, d \mu \geq \int_E a \, d \mu = a \int_E \, d \mu = a \mu(E)
\end{equation}
and
\begin{equation}
\int_E f \, d \mu \leq \int_E b \, d \mu = b \int_E \, d \mu = b \mu(E)
\end{equation}
The latter two results along with result
R1902: Signed basic integral of almost everywhere equal functions also establish claims (2) and (3). $\square$