Proof of Gronwall's Lemma for Bounded Measurable Functions

Choose $K\ge0$ with $|u(s)|\le K$ for all $s\in[0,T]$. Since $u$ is measurable and bounded and $[0,T]$ carries a finite restricted Lebesgue measure, $u$ is integrable on $[0,t]$ for every $t\in[0,T]$, so the hypothesis is meaningful.

Define $v:[0,T]\to\mathbb{R}$ by

Step 1: $v$ is continuous. Let $0\le r\le t\le T$. Writing $u\mathbf{1}_{[0,t]}=u\mathbf{1}_{[0,r]}+u\mathbf{1}_{(r,t]}$, where $\mathbf{1}_D$ denotes the function equal to $1$ on $D$ and $0$ elsewhere, and using linearity of the integral,

Since $-K\le u\le K$, monotonicity of the integral gives $|v(t)-v(r)|\le cK\,(t-r)$. Hence, given $\varepsilon>0$, the choice $\delta=\varepsilon/(cK+1)>0$ makes $|v(t)-v(r)|<\varepsilon$ whenever $t,r\in[0,T]$ satisfy $|t-r|<\delta$, so $v$ is continuous on $[0,T]$, where both the domain $[0,T]$ and the codomain $\mathbb{R}$ carry the metric of the real line. In particular $v$ is measurable by claim 4 of Measurability of Countable Suprema, Bounded Pointwise Limits, Monotone Functions, and Continuous Functions, and $v$ is bounded, since $|v(t)|\le|a|+cKT$ for every $t\in[0,T]$, so it is integrable on each $[0,t]$.

Step 2: $v$ satisfies the same inequality. By hypothesis $u(t)\le v(t)$ for every $t\in[0,T]$. Since $c\ge0$, monotonicity of the integral gives, for every $t\in[0,T]$,

Step 3: conclusion. Let $t\in(0,T]$. The restriction of $v$ to $[0,t]$ is continuous by claim 1 of Restriction Stability of Continuity and of the Derivative, so by claim 3 of Restricted Lebesgue Measure and Integral Toolkit on a Compact Interval that restriction is Riemann integrable on $[0,t]$ and its Lebesgue integral over $[0,t]$ equals its Riemann integral there; for $t=0$ both integrals are $0$ by convention. Thus the inequality of Step 2 is exactly the hypothesis of Gronwall's lemma in integral form, applied to the continuous function $v$ on $[0,T]$ with the constants $a$ and $c\ge0$, and that lemma yields

Combining with $u(t)\le v(t)$ gives $u(t)\le a\exp(ct)$ for every $t\in[0,T]$. $\blacksquare$