= Solution
Write $\theta_0=\theta^{(t)}$ and $h_0(z)=f(x,z;\theta_0)/L(x\mid\theta_0)$. Assume the likelihoods are positive and the expected log likelihoods used in the step are finite. On the support where $h_0>0$, let $r(z)=f(x,z;\theta)/f(x,z;\theta_0)$. Integrating over that support gives
$$
\frac{L(x\mid\theta)}{L(x\mid\theta_0)}\ge\int h_0(z)r(z)\,dz.
$$
It is an equality when the candidate joint density has no additional mass outside the old support. Concavity of the logarithm and the <Jensen inequality> now imply
$$
\begin{aligned}
\log L(x\mid\theta)-\log L(x\mid\theta_0)&\ge\log\mathbb E_{h_0}r(Z)\\
&\ge\mathbb E_{h_0}\log r(Z)\\
&=Q(\theta\mid\theta_0)-Q(\theta_0\mid\theta_0).
\end{aligned}
$$
Thus the argument does not silently require an identical support at every parameter. The M-step selects a value whose right-hand side is nonnegative, proving <EM likelihood monotonicity>:
$$
\boxed{\log L(x\mid\theta^{(t+1)})\ge\log L(x\mid\theta^{(t)}).}
$$
When the conditional densities allow finite divergence, the same calculation has the informative exact form
$$
\log L(x\mid\theta)-\log L(x\mid\theta_0)=Q(\theta\mid\theta_0)-Q(\theta_0\mid\theta_0)+\operatorname{KL}(h_0\Vert h_\theta).
$$
The <Kullback-Leibler divergence> is nonnegative, since $\mathbb E_{h_0}\log(h_\theta/h_0)\le\log\int_{\{h_0>0\}}h_\theta\le0$. This also explains why an increase of the surrogate objective guarantees an increase of the observed objective, while equality of the two likelihoods is possible.
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