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Q:

How to show $\int_{0}^{\infty} f(x) dx = \sum_{i=0}^{\infty} \alpha_i$ in book?

Given a real-valued function $f$ that is continuous on $(0,\infty)$ and $(0,1]$, how to show that $$\int_{0}^{\infty} f(x) dx = \sum_{i=0}^{\infty} \alpha_i$$
is true in a series sense?

Use the definition of Lebesgue integral. Since $f$ is continuous on $(0,\infty)$ and $(0,1]$, this part works out.

y|\psi_k)$ and then applying a minimization to the remaining variables. The first infidelities per basis are $$\begin{aligned}
\varepsilon_{c_l} &=& \max_{y^N \in \operatorname{Range}(\mathcal{M})} \min_{x^l} |f(y^N||c_l) – \sum_{k=1}^K y^k \operatorname*{Tr}[\mathcal{N}_k \mathcal{M}_k]|\\
\varepsilon_{u_l} &=& \max_{y^N \in \operatorname{Range}(\mathcal{M})} \min_{x^l} |f(y^N||u_l) – \sum_{k=1}^K y^k \operatorname*{Tr}[\mathcal{M}_k \mathcal{N}_k]|\end{aligned}$$ For $u_l$ to be infidel, the maximum infidelity over range of all measurements will be $$\varepsilon = \max_{y^N \in \operatorname{Range}(\mathcal{M})} \min_{x^l,x^r} |f(y^N||u_l) – \sum_{k=1}^K y^k
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