= Solution
The <strong law of large numbers> gives
$$
\boxed{\widehat\mu_n=\frac1n\sum_{i=1}^nh(X_i)
\longrightarrow\mathbb E_{\theta_0}[h(X)]=\mu(\theta_0)
\quad\text{almost surely}.}
$$
A continuous one-to-one real function is strictly monotone and has a continuous inverse on its image. Hence
$$
\boxed{\widehat\theta_n=\mu^{-1}(\widehat\mu_n)}
$$
whenever $\widehat\mu_n$ lies in that image; an arbitrary extension outside the image handles the remaining finite-sample values. The <continuous mapping theorem> gives $\widehat\theta_n\to\theta_0$ almost surely, so this is a <consistent estimator>.
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