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 $[0,T]$ has finite Lebesgue measure, $u$ is integrable on $[0,t]$ for every $t$, so the hypothesis is meaningful.

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

$$v(t)=a+c\int_{[0,t]}u(s)\,ds .$$

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,

$$v(t)-v(r)=c\int_{(r,t]}u(s)\,ds .$$

Since $-K\le u\le K$, monotonicity of the integral gives $|v(t)-v(r)|\le cK\,(t-r)$. Hence $v$ is continuous on $[0,T]$; in particular $v$ is measurable and bounded, 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 $t\in[0,T]$,

$$v(t)=a+c\int_{[0,t]}u(s)\,ds\le a+c\int_{[0,t]}v(s)\,ds .$$

Step 3: conclusion. The function $v$ is continuous, so its Lebesgue integral over $[0,t]$ coincides with its Riemann integral there, by the compact-interval toolkit. Thus the inequality of Step 2 is exactly the hypothesis of Gronwall's lemma in integral form for the continuous function $v$ with constants $a$ and $c\ge0$, and that lemma yields

$$v(t)\le a\,e^{ct}\qquad\text{for every }t\in[0,T].$$

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