It looks to me like the authors just wanted to put in some ridiculous values which "are clearly sufficient" so that they don't have to work out the details.

The choice $\eta = c_1^{1/10}$ should ensure that the term $\int_{|g-1|\leq \eta} |g|^\beta sgn(g) d\mu$ converges to 1 for $c_1 \rightarrow 0$, uniformly across all $g$ satisfying $\mathbb{E}[g^2] = 1$ and $\mathbb{E}[(g-\mathbb{E}[g])^2 \leq c_1$ and uniformly across all $\beta$ (this is where you need an upper bound on $\beta$). The term $\sqrt{3/4}$ is then just some value that is sufficiently close to 1 for the remainder of the proof.

To see this convergence, it suffices to show that $\mu(|g-1|\leq \eta)$ goes to 1 using, e.g., Chebyshev's inequality. To this end, we need to use the vague statement "$|1-\mathbb{E}[g]| \ll 1$", which should be made rigorous by establishing a bound on $|\mathbb{E}[g] - 1|$ only depending on $c_1$. I haven't worked out the details, but I guess something like $|\mathbb{E}[g] - 1| \leq f(c_1)$, where $f(x) \approx x^{1/2}$ seems reasonable. The choice of $\eta$ is now simply large enough such that \begin{align} \mu(|g-1| > \eta) \leq \mu(|g-\mathbb{E}[g]| > \eta - f(c_1)) \leq \mu(|g-\mathbb{E}[g]| > c_1^{1/3}) \rightarrow 0 \end{align} for $c_1 \rightarrow 0$, where of course the $1/3$ exponent is again a rather arbitrary choice by myself, as anything below $1/2$ should work.

It looks to me like the authors just wanted to put in some ridiculous values which "are clearly sufficient" so that they don't have to work out the details.

The choice $\eta = c_1^{1/10}$ should ensure that the term $\int_{|g-1|\leq \eta} |g|^\beta \operatorname{sgn}(g) \, d\mu$ converges to 1 for $c_1 \rightarrow 0$, uniformly across all $g$ satisfying $\mathbb{E}[g^2] = 1$ and $\mathbb{E}[(g-\mathbb{E}[g])^2 \leq c_1$ and uniformly across all $\beta$ (this is where you need an upper bound on $\beta$). The term $\sqrt{3/4}$ is then just some value that is sufficiently close to 1 for the remainder of the proof.

To see this convergence, it suffices to show that $\mu(|g-1|\leq \eta)$ goes to 1 using, e.g., Chebyshev's inequality. To this end, we need to use the vague statement "$|1-\mathbb{E}[g]| \ll 1$", which should be made rigorous by establishing a bound on $|\mathbb{E}[g] - 1|$ only depending on $c_1$. I haven't worked out the details, but I guess something like $|\mathbb{E}[g] - 1| \leq f(c_1)$, where $f(x) \approx x^{1/2}$ seems reasonable. The choice of $\eta$ is now simply large enough such that \begin{align} \mu(|g-1| > \eta) \leq \mu(|g-\mathbb{E}[g]| > \eta - f(c_1)) \leq \mu(|g-\mathbb{E}[g]| > c_1^{1/3}) \rightarrow 0 \end{align} for $c_1 \rightarrow 0$, where of course the $1/3$ exponent is again a rather arbitrary choice by myself, as anything below $1/2$ should work.

It looks to me like the authors just wanted to put in some ridiculous values which "are clearly sufficient" so that they don't have to work out the details.

The choice $\eta = c_1^{1/10}$ should ensure that the term $\int_{|g-1|\leq \eta} |g|^\beta sgn(g) d\mu$ converges to 1 for $c_1 \rightarrow 0$, uniformly across all $g$ satisfying $\mathbb{E}[g^2] = 1$ and $\mathbb{E}[(g-\mathbb{E}[g])^2 \leq c_1$ and uniformly across all $\beta$ (this is where you need an upper bound on $\beta$). The term $\sqrt{3/4}$ is then just some value that is sufficiently close to 1 for the remainder of the proof.

To see this convergence, it suffices to show that $\mu(|g-1|\leq \eta)$ goes to 1 using, e.g., Chebyshev's inequality. To this end, we need to use the vague statement "$|1-\mathbb{E}[g]| \ll 1$", which should be made rigorous by establishing a bound on $|\mathbb{E}[g] - 1|$ only depending on $c_1$. I haven't worked out the details, but I guess something like $|\mathbb{E}[g] - 1| \leq f(c_1)$, where $f(x) \approx x^{1/2}$ seems reasonable. The choice of $\eta$ is now simply large enough such that \begin{align} \mu(|g-1| > \eta) \leq \mu(|g-\mathbb{E}[g]| > \eta - f(c_1)) \leq \mu(|g-\mathbb{E}[g]| > c_1) \rightarrow 0 \end{align} for $c_1 \rightarrow 0$.

It looks to me like the authors just wanted to put in some ridiculous values which "are clearly sufficient" so that they don't have to work out the details.

The choice $\eta = c_1^{1/10}$ should ensure that the term $\int_{|g-1|\leq \eta} |g|^\beta sgn(g) d\mu$ converges to 1 for $c_1 \rightarrow 0$, uniformly across all $g$ satisfying $\mathbb{E}[g^2] = 1$ and $\mathbb{E}[(g-\mathbb{E}[g])^2 \leq c_1$ and uniformly across all $\beta$ (this is where you need an upper bound on $\beta$). The term $\sqrt{3/4}$ is then just some value that is sufficiently close to 1 for the remainder of the proof.

To see this convergence, it suffices to show that $\mu(|g-1|\leq \eta)$ goes to 1 using, e.g., Chebyshev's inequality. To this end, we need to use the vague statement "$|1-\mathbb{E}[g]| \ll 1$", which should be made rigorous by establishing a bound on $|\mathbb{E}[g] - 1|$ only depending on $c_1$. I haven't worked out the details, but I guess something like $|\mathbb{E}[g] - 1| \leq f(c_1)$, where $f(x) \approx x^{1/2}$ seems reasonable. The choice of $\eta$ is now simply large enough such that \begin{align} \mu(|g-1| > \eta) \leq \mu(|g-\mathbb{E}[g]| > \eta - f(c_1)) \leq \mu(|g-\mathbb{E}[g]| > c_1^{1/3}) \rightarrow 0 \end{align} for $c_1 \rightarrow 0$, where of course the $1/3$ exponent is again a rather arbitrary choice by myself, as anything below $1/2$ should work.