for $t$ with $\sigma_i(t) < 1$ and $\theta_i(t) < 1$. Hence, $d_\mu(t) \leq \varepsilon_\text{r} \leq \varepsilon$ holds for $t$ with $\sigma(t) \geq 1$ or $\theta(t) \geq 1$, and $\sigma(t) \leq m-1$ or $\theta(t) \leq n -1$.

To show that $\mu$ is locally correct, suppose for contradiction that values $a,b$ exist such that $\delta_\text{F}(P_\sigma[a,b], Q_\theta[a,b]) < d_\mu[a,b]$. If $a,b$ are in between two consecutive events, we know that the submatching corresponds to one of the matchings $\mu_i$. Since these are locally correct, $\delta_\text{F}(P_\sigma[a,b], Q_\theta[a,b]) = d_\mu[a,b]$ must hold.

Hence, suppose that $a$ and $b$ are separated by at least one event of $E_\text{r}$. There are two possibilities: either $d_\mu[a,b] = \varepsilon_\text{r}$ or $d_\mu[a,b] > \varepsilon_\text{r}$. $d_\mu[a,b] < \varepsilon_\text{r}$ cannot hold, since $d_\mu[a,b]$ includes a realizing event. First, assume $d_\mu[a,b] = \varepsilon_\text{r}$ holds. If $\delta_\text{F}(P_\sigma[a,b], Q_\theta[a,b]) < \varepsilon_\text{r}$ holds, then a matching exists that does not use the events between $a$ and $b$ and has a lower maximum. Hence, the free space connects point $(\sigma(a),\theta(a))$ with point $(\sigma(b),\theta(b))$ at a lower value than $\varepsilon_\text{r}$. This implies that all events between $a$ and $b$ can be omitted, contradicting that $E_\text{r}$ is a minimal realizing set.

Now, assume $d_\mu[a,b] > \varepsilon_\text{r}$. Let $t'$ denote 