[proofplan]
We first apply entropy tensorization to the product structure generated by the independent coordinates $X_1,\dots,X_n$. Each conditional entropy is bounded by comparing $F$ with the auxiliary coordinate-deleted value $F_i$ and using the one-sided increment bound $0\le F-F_i\le 1$. Summing these local estimates and using the self-bounding inequality gives the entropy estimate. We then convert the entropy estimate into a differential inequality for the centered log-[Laplace transform](/page/Laplace%20Transform) and integrate it by the Herbst argument. Finally, Markov's inequality and an explicit optimization in the Laplace parameter give the stated upper tail bound.
[/proofplan]
[step:Define the coordinate increments and conditional entropy operators]
For $i\in\{1,\dots,n\}$, define $X_{-i}:=(X_1,\dots,X_{i-1},X_{i+1},\dots,X_n)$ and write $E_{-i}:=E_1\times\cdots\times E_{i-1}\times E_{i+1}\times\cdots\times E_n$. By the self-bounding hypothesis, there is a measurable coordinate-deleted map
\begin{align*}
F_i:E_{-i}\to[0,\infty)
\end{align*}
Thus $F_i(X_{-i})$ is a [random variable](/page/Random%20Variable) depending on all coordinates except $X_i$. Define the increment random variable
\begin{align*}
\Delta_i:\Omega\to[0,1],\qquad \Delta_i:=F(X)-F_i(X_{-i})
\end{align*}
The self-bounding hypotheses say that
\begin{align*}
0\le \Delta_i\le 1
\end{align*}
for each $i$, and
\begin{align*}
\sum_{i=1}^n\Delta_i\le F(X).
\end{align*}
Let $\mathcal L^1$ denote one-dimensional [Lebesgue measure](/page/Lebesgue%20Measure). Let $\mu_i$ denote the law of $X_i$ on $(E_i,\mathcal E_i)$. For a non-negative random variable $Y=Y(X_1,\dots,X_n)$ with finite expectations in the following expressions and a coordinate $i$, define $\mathbb E_i[Y]$ to be the function of $X_{-i}$ obtained by integrating only the $i$-th coordinate against $\mu_i$:
\begin{align*}
\mathbb E_i[Y](X_{-i}) := \int_{E_i} Y(X_1,\dots,X_{i-1},x_i,X_{i+1},\dots,X_n)\,d\mu_i(x_i).
\end{align*}
This definition uses the product structure coming from independence and does not require a regular [conditional probability](/page/Conditional%20Probability) kernel. Define the corresponding conditional entropy by
\begin{align*}
\operatorname{Ent}_i(Y):=\mathbb E_i[Y\log Y]-\mathbb E_i[Y]\log \mathbb E_i[Y].
\end{align*}
We use the entropy tensorization inequality for product measures:
\begin{align*}
\operatorname{Ent}(Y)\le \mathbb E\left[\sum_{i=1}^n \operatorname{Ent}_i(Y)\right].
\end{align*}
Here the structural hypothesis needed for tensorization is exactly the independence of $X_1,\dots,X_n$; the displayed inequality is applied only when the global and conditional entropy terms are finite.
[/step]
[step:Bound each conditional entropy by the coordinate increment]
Fix $\lambda>0$ and define
\begin{align*}
Y:\Omega\to[0,\infty),\qquad Y:=e^{\lambda F(X)}
\end{align*}
For each $i$, condition on $X_{-i}$. Under this conditioning, $F_i(X_{-i})$ is constant as a function of $X_i$. The variational entropy bound gives, for every positive constant $a$,
\begin{align*}
\operatorname{Ent}_i(Y)\le \mathbb E_i\left[Y\log\frac{Y}{a}-Y+a\right].
\end{align*}
Choose
\begin{align*}
a:=e^{\lambda F_i(X_{-i})}.
\end{align*}
Since $Y=e^{\lambda F(X)}$ and $\Delta_i=F(X)-F_i(X_{-i})$, this gives
\begin{align*}
\operatorname{Ent}_i(e^{\lambda F(X)})
\le \mathbb E_i\left[e^{\lambda F(X)}\left(\lambda\Delta_i-1+e^{-\lambda\Delta_i}\right)\right].
\end{align*}
For $u\ge 0$,
\begin{align*}
e^{-u}+u-1=\int_0^u (u-s)e^{-s}\,d\mathcal L^1(s)\le \frac{u^2}{2}.
\end{align*}
Applying this with $u=\lambda\Delta_i$ and using $0\le\Delta_i\le1$, we obtain
\begin{align*}
e^{-\lambda\Delta_i}+\lambda\Delta_i-1\le \lambda^2 e^\lambda \Delta_i.
\end{align*}
Therefore
\begin{align*}
\operatorname{Ent}_i(e^{\lambda F(X)})
\le \lambda^2 e^\lambda\,\mathbb E_i[\Delta_i e^{\lambda F(X)}].
\end{align*}
[guided]
Fix one coordinate $i$. The goal is to estimate the entropy coming only from the randomness of $X_i$, while all other coordinates are frozen. Under this conditioning, the quantity $F_i(X_{-i})$ is constant because it does not depend on $X_i$.
We apply the variational entropy bound
\begin{align*}
\operatorname{Ent}_i(Y)\le \mathbb E_i\left[Y\log\frac{Y}{a}-Y+a\right],
\end{align*}
valid for every positive constant $a$ with respect to integration against the marginal law $\mu_i$ of $X_i$. We use it with
\begin{align*}
Y:=e^{\lambda F(X)}
\end{align*}
and choose the comparison constant
\begin{align*}
a:=e^{\lambda F_i(X_{-i})}.
\end{align*}
This is the natural comparison because the difference between $F(X)$ and $F_i(X_{-i})$ is precisely the controlled coordinate increment $\Delta_i$.
Since
\begin{align*}
\log\frac{e^{\lambda F(X)}}{e^{\lambda F_i(X_{-i})}}=\lambda(F(X)-F_i(X_{-i}))=\lambda\Delta_i
\end{align*}
and
\begin{align*}
e^{\lambda F_i(X_{-i})}=e^{\lambda F(X)}e^{-\lambda\Delta_i},
\end{align*}
the variational bound becomes
\begin{align*}
\operatorname{Ent}_i(e^{\lambda F(X)})
\le \mathbb E_i\left[e^{\lambda F(X)}\left(\lambda\Delta_i-1+e^{-\lambda\Delta_i}\right)\right].
\end{align*}
It remains to estimate the scalar factor in parentheses. For $u\ge0$, Taylor's formula with integral remainder gives
\begin{align*}
e^{-u}+u-1=\int_0^u (u-s)e^{-s}\,d\mathcal L^1(s).
\end{align*}
Since $e^{-s}\le1$ for $s\ge0$,
\begin{align*}
e^{-u}+u-1\le \int_0^u (u-s)\,d\mathcal L^1(s)=\frac{u^2}{2}.
\end{align*}
Now set $u=\lambda\Delta_i$. The self-bounding hypothesis gives $0\le\Delta_i\le1$, hence $\Delta_i^2\le\Delta_i$. Therefore
\begin{align*}
e^{-\lambda\Delta_i}+\lambda\Delta_i-1\le \frac{\lambda^2\Delta_i^2}{2}\le \lambda^2 e^\lambda\Delta_i.
\end{align*}
Substituting this into the conditional entropy estimate yields
\begin{align*}
\operatorname{Ent}_i(e^{\lambda F(X)})
\le \lambda^2 e^\lambda\,\mathbb E_i[\Delta_i e^{\lambda F(X)}].
\end{align*}
[/guided]
[/step]
[step:Sum the local estimates to prove the entropy inequality]
Apply entropy tensorization to $Y=e^{\lambda F(X)}$ and then use the conditional estimate from the previous step:
\begin{align*}
\operatorname{Ent}(e^{\lambda F(X)})
\le \mathbb E\left[\sum_{i=1}^n \operatorname{Ent}_i(e^{\lambda F(X)})\right].
\end{align*}
Thus
\begin{align*}
\operatorname{Ent}(e^{\lambda F(X)})
\le \lambda^2 e^\lambda\,\mathbb E\left[e^{\lambda F(X)}\sum_{i=1}^n\Delta_i\right].
\end{align*}
Using the self-bounding inequality $\sum_{i=1}^n\Delta_i\le F(X)$, we obtain
\begin{align*}
\operatorname{Ent}(e^{\lambda F(X)})
\le \lambda^2 e^\lambda\,\mathbb E[F(X)e^{\lambda F(X)}].
\end{align*}
This proves the asserted entropy estimate.
[/step]
[step:Convert the entropy estimate into a differential inequality for the log-Laplace transform]
Let
\begin{align*}
m:=\mathbb E[F(X)].
\end{align*}
If $m=0$, then $F(X)=0$ $\mathbb P$-a.s. because $F(X)\ge0$, and both the entropy estimate and the tail bound are immediate. Hence assume $m>0$ for the Laplace-transform argument.
For the concentration conclusion, assume
\begin{align*}
\mathbb E[e^{aF(X)}]<\infty
\end{align*}
for every $0<a<1/e$. For every $0<\lambda<1/e$, choose $\varepsilon>0$ such that $\lambda+\varepsilon<1/e$. Then $F(X)e^{\lambda F(X)}\le \varepsilon^{-1}e^{(\lambda+\varepsilon)F(X)}$, so all differentiations below are justified by dominated convergence. Define the centered log-Laplace transform on $(0,1/e)$ by
\begin{align*}
\psi:(0,1/e)\to\mathbb R,
\lambda\mapsto\log\mathbb E[e^{\lambda(F(X)-m)}]
\end{align*}
For such $\lambda$,
\begin{align*}
\mathbb E[e^{\lambda F(X)}]=e^{\lambda m+\psi(\lambda)}.
\end{align*}
Therefore differentiating under the expectation gives
\begin{align*}
\frac{\mathbb E[F(X)e^{\lambda F(X)}]}{\mathbb E[e^{\lambda F(X)}]}=m+\psi'(\lambda).
\end{align*}
The entropy identity for $e^{\lambda F(X)}$ is
\begin{align*}
\operatorname{Ent}(e^{\lambda F(X)})
=\mathbb E[e^{\lambda F(X)}]\left(\lambda\psi'(\lambda)-\psi(\lambda)\right).
\end{align*}
Dividing the entropy estimate by the positive number $\mathbb E[e^{\lambda F(X)}]$ gives
\begin{align*}
\lambda\psi'(\lambda)-\psi(\lambda)
\le \lambda^2 e^\lambda\left(m+\psi'(\lambda)\right).
\end{align*}
For $0<\lambda<1/e$, we have $e^\lambda\le e$, so
\begin{align*}
\lambda\psi'(\lambda)-\psi(\lambda)
\le e\lambda^2\left(m+\psi'(\lambda)\right).
\end{align*}
[/step]
[step:Integrate the differential inequality by the Herbst argument]
We now carry out the Herbst argument. For $0<\lambda<1/e$, define
\begin{align*}
h:(0,1/e)\to\mathbb R,
\lambda\mapsto\frac{\psi(\lambda)}{\lambda}
\end{align*}
Since $\psi(0)=0$ and $\psi'(0)=\mathbb E[F(X)-m]=0$, we have $\lim_{\lambda\downarrow0}h(\lambda)=0$. Also,
\begin{align*}
h'(\lambda)=\frac{\lambda\psi'(\lambda)-\psi(\lambda)}{\lambda^2}.
\end{align*}
The differential inequality from the previous step gives
\begin{align*}
h'(\lambda)\le e(m+\psi'(\lambda)).
\end{align*}
Because $\psi(\lambda)=\lambda h(\lambda)$, we have
\begin{align*}
\psi'(\lambda)=h(\lambda)+\lambda h'(\lambda).
\end{align*}
Therefore
\begin{align*}
h'(\lambda)\le e(m+h(\lambda)+\lambda h'(\lambda)).
\end{align*}
Rearranging, and using $1-e\lambda>0$, yields
\begin{align*}
\frac{d}{d\lambda}\log(m+h(\lambda))\le \frac{e}{1-e\lambda}.
\end{align*}
Integrating over $(0,\lambda)$ with respect to $\mathcal L^1$ gives
\begin{align*}
\log\frac{m+h(\lambda)}{m}\le -\log(1-e\lambda).
\end{align*}
Hence
\begin{align*}
h(\lambda)\le \frac{e\lambda m}{1-e\lambda}.
\end{align*}
Multiplying by $\lambda$ gives
\begin{align*}
\psi(\lambda)\le \frac{e\lambda^2 m}{1-e\lambda}
\end{align*}
for every $0<\lambda<1/e$.
[/step]
[step:Apply Markov's inequality and optimize the Laplace parameter]
Fix $t>0$ and $0<\lambda<1/e$. Markov's inequality applied to the non-negative random variable $e^{\lambda(F(X)-m)}$ gives
\begin{align*}
\mathbb P(F(X)-m\ge t)
\le \exp(-\lambda t+\psi(\lambda)).
\end{align*}
Using the Laplace estimate from the previous step,
\begin{align*}
\mathbb P(F(X)-m\ge t)
\le \exp\left(-\lambda t+\frac{e\lambda^2 m}{1-e\lambda}\right).
\end{align*}
Choose
\begin{align*}
\lambda:=\frac{t}{e(2m+t)}.
\end{align*}
Then $0<\lambda<1/e$ and
\begin{align*}
1-e\lambda=1-\frac{t}{2m+t}=\frac{2m}{2m+t}.
\end{align*}
Substituting this value into the exponent gives
\begin{align*}
-\lambda t+\frac{e\lambda^2 m}{1-e\lambda}
= -\frac{t^2}{e(2m+t)}+\frac{t^2}{2e(2m+t)}
= -\frac{t^2}{2e(2m+t)}.
\end{align*}
Since $m=\mathbb E[F(X)]$, this is
\begin{align*}
\mathbb P(F(X)-\mathbb E[F(X)]\ge t)
\le \exp\left(-\frac{t^2}{4e\,\mathbb E[F(X)]+2e\,t}\right).
\end{align*}
This proves the displayed concentration inequality, with the denominator $4e\,\mathbb E[F(X)]+2e\,t$ obtained from the entropy estimate proved above.
[/step]
