= Solution
Fix the observed $Y$, let $L(\theta)=p_\theta(Y)$, and let $q_0(z)=p_{\theta_0}(z\mid Y)$. The E-step forms
$$
Q(\theta\mid\theta_0)=\int q_0(z)\log p_\theta(Y,z)\,dz.
$$
For finite expected log-likelihoods and $L(\theta_0)>0$, the <Jensen inequality> gives
$$
\begin{aligned}
\log\frac{L(\theta)}{L(\theta_0)}
&\geq\log\int q_0(z)\frac{p_\theta(Y,z)}{p_{\theta_0}(Y,z)}\,dz\\
&\geq\int q_0(z)\log\frac{p_\theta(Y,z)}{p_{\theta_0}(Y,z)}\,dz\\
&=Q(\theta\mid\theta_0)-Q(\theta_0\mid\theta_0).
\end{aligned}
$$
The first inequality allows additional mass outside the old complete-data support; it is equality when the relevant supports coincide. Ratios are taken on the support of $q_0$. If a candidate assigns zero density on positive $q_0$-mass, its $Q$ is minus infinity and it cannot be an improving finite M-step.
An M-step of the <expectation-maximization algorithm> chooses $\theta_1$ with $Q(\theta_1\mid\theta_0)\geq Q(\theta_0\mid\theta_0)$, so
$$
\boxed{\log L(\theta_1)\geq\log L(\theta_0),\qquad L(\theta_1)\geq L(\theta_0).}
$$
This <EM likelihood monotonicity> also holds for a generalized M-step that merely increases $Q$, rather than maximizing it exactly. It is a statement about the observed-data <likelihood function>.
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