[proofplan]
We argue by contradiction. If $g(t)$ does not tend to $0$, then there are arbitrarily large times at which $g$ is bounded below by one fixed positive number. [Uniform continuity](/page/Uniform%20Continuity) turns each such peak into an interval of fixed positive length on which $g$ is still bounded below. By choosing the peak times far apart, these intervals are disjoint, and their fixed positive contributions force the integral of $g$ over $[0,\infty)$ to be infinite, contradicting the hypothesis.
[/proofplan]
[step:Extract arbitrarily large points where $g$ stays uniformly positive]
Assume, for contradiction, that $\lim_{t \to \infty} g(t) \ne 0$. Since $g(t) \geq 0$ for every $t \in [0,\infty)$, this means that there exists a number $\varepsilon > 0$ such that for every $R > 0$ there is a point $t \in [R,\infty)$ with
\begin{align*}
g(t) \geq \varepsilon.
\end{align*}
Indeed, if no such $\varepsilon$ existed, then for every $\varepsilon > 0$ there would be $R_\varepsilon > 0$ such that $0 \leq g(t) < \varepsilon$ for all $t \geq R_\varepsilon$, which is exactly $\lim_{t \to \infty} g(t) = 0$.
[/step]
[step:Use uniform continuity to make every peak into a fixed-width lower bound]
By uniform continuity of $g$, applied with the tolerance $\varepsilon/2 > 0$, there exists $\delta > 0$ such that for all $s,t \in [0,\infty)$,
\begin{align*}
|s-t| < \delta \implies |g(s)-g(t)| < \frac{\varepsilon}{2}.
\end{align*}
We will use the interval radius $r := \delta/2 > 0$. Whenever $t \in [0,\infty)$ satisfies $g(t) \geq \varepsilon$ and $s \in [0,\infty)$ satisfies $|s-t| < r$, we have $|s-t| < \delta$, hence
\begin{align*}
g(s) \geq g(t) - |g(s)-g(t)| > \varepsilon - \frac{\varepsilon}{2} = \frac{\varepsilon}{2}.
\end{align*}
[guided]
The purpose of uniform continuity is to make the same interval size work around every large peak. We apply uniform continuity to the positive tolerance $\varepsilon/2$. Thus there is a number $\delta > 0$ such that for all $s,t \in [0,\infty)$,
\begin{align*}
|s-t| < \delta \implies |g(s)-g(t)| < \frac{\varepsilon}{2}.
\end{align*}
Define the radius $r := \delta/2$. This choice ensures that $|s-t| < r$ is certainly strong enough to imply $|s-t| < \delta$. Now suppose $t \in [0,\infty)$ is one of the points where $g(t) \geq \varepsilon$, and suppose $s \in [0,\infty)$ lies within distance $r$ of $t$. Then uniform continuity gives
\begin{align*}
|g(s)-g(t)| < \frac{\varepsilon}{2}.
\end{align*}
Taking the lower bound direction from this absolute-value inequality,
\begin{align*}
g(s) \geq g(t) - |g(s)-g(t)|.
\end{align*}
Since $g(t) \geq \varepsilon$, we obtain
\begin{align*}
g(s) > \varepsilon - \frac{\varepsilon}{2} = \frac{\varepsilon}{2}.
\end{align*}
Thus every point where $g$ reaches height at least $\varepsilon$ produces an interval of radius $r$ on which $g$ is bounded below by $\varepsilon/2$.
[/guided]
[/step]
[step:Choose the peak intervals disjoint and contained in $[0,\infty)$]
Let $\mathbb{N} := \{1,2,3,\dots\}$ denote the set of positive integers. Using the arbitrarily large positive peaks from the first step, choose a sequence $(t_k)_{k=1}^{\infty}$ in $[0,\infty)$ recursively so that
\begin{align*}
g(t_k) \geq \varepsilon
\end{align*}
and
\begin{align*}
t_1 > r, \qquad t_{k+1} > t_k + 2r \quad \text{for every } k \in \mathbb{N}.
\end{align*}
For each $k \in \mathbb{N}$, define the open interval
\begin{align*}
I_k := (t_k-r,t_k+r) \subset [0,\infty).
\end{align*}
The condition $t_1 > r$ ensures $I_1 \subset [0,\infty)$, and the recursive inequality ensures that the intervals $I_k$ are pairwise disjoint. Each interval has one-dimensional [Lebesgue measure](/page/Lebesgue%20Measure)
\begin{align*}
\mathcal{L}^1(I_k) = 2r = \delta.
\end{align*}
[/step]
[step:Sum the fixed positive contributions to contradict integrability]
Let
\begin{align*}
M := \int_{[0,\infty)} g(t)\,d\mathcal{L}^1(t).
\end{align*}
The hypothesis that $g$ is Lebesgue integrable supplies measurability of $g$ with respect to $\mathcal{L}^1$ and gives $M < \infty$. For every $k \in \mathbb{N}$ and every $s \in I_k$, the preceding lower-bound step gives $g(s) > \varepsilon/2$. Hence, for each $N \in \mathbb{N}$, monotonicity and finite additivity of the [Lebesgue integral](/page/Lebesgue%20Integral) over the pairwise disjoint measurable intervals $I_1,\dots,I_N$ give
\begin{align*}
M \geq \int_{\bigcup_{k=1}^{N} I_k} g(t)\,d\mathcal{L}^1(t).
\end{align*}
Since $g(t) > \varepsilon/2$ on each $I_k$,
\begin{align*}
\int_{\bigcup_{k=1}^{N} I_k} g(t)\,d\mathcal{L}^1(t) \geq \sum_{k=1}^{N} \int_{I_k} \frac{\varepsilon}{2}\,d\mathcal{L}^1(t).
\end{align*}
Using $\mathcal{L}^1(I_k)=\delta$ for every $k$,
\begin{align*}
\sum_{k=1}^{N} \int_{I_k} \frac{\varepsilon}{2}\,d\mathcal{L}^1(t) = \sum_{k=1}^{N} \frac{\varepsilon}{2}\mathcal{L}^1(I_k) = N\frac{\varepsilon\delta}{2}.
\end{align*}
Therefore
\begin{align*}
M \geq N\frac{\varepsilon\delta}{2}
\end{align*}
for every $N \in \mathbb{N}$. This is impossible because $M < \infty$ while $N\varepsilon\delta/2 \to \infty$ as $N \to \infty$. The contradiction proves that
\begin{align*}
\lim_{t \to \infty} g(t) = 0.
\end{align*}
[/step]
