 Solution

For a fixed $\lambda <{\lambda }_0$, the likelihood ratio against the null isstrictly increasing in $S={\sum }_i{x}_i$. The Neyman–Pearson critical region is therefore an upper tail in $S$, and its null threshold depends only on the desired size, not on which $\lambda <{\lambda }_0$ was chosen. Thus the same test is uniformly most powerful for the entire one-sided alternative.

Under the null $S\sim \mathrm{Gamma}(n,{\lambda }_0)$ with the second parameter interpreted as rate. Its continuous distribution permits the exact thresholdUnder any fixed alternative its rejection probability is $\boxed{Q(\lambda )=1-F(c_{\alpha };n,\lambda )}$. Applying the lemma separately to each such $\lambda$ shows that every level-$\alpha$ test has power at most this value there. The constructed test attains the bound for all $\lambda <{\lambda }_0$ simultaneously.