Proof of Gronwall's Lemma for Bounded Measurable Functions

The idea is to pass from $u$ to the continuous majorant given by the right-hand side of the hypothesis, to which the continuous form of Gronwall's lemma applies.

Step 1: a continuous majorant. Let $w:[a,b]\to\mathbb{R}$ be the function

$$w(t)=\alpha+\beta\int_{[a,t]}u\,d\lambda\qquad(t\in[a,b]).$$

The indefinite integral $t\mapsto\int_{[a,t]}u\,d\lambda$, a function on $[a,b]$ by Functions and Function-Valued Operators Given by Integrals over Subintervals (i) as $u\in\mathcal{L}^{1}([a,b])$, is continuous on $[a,b]$ (Indefinite Lebesgue Integrals on a Closed Interval (v)), hence so is $w$, a constant plus a constant multiple of it (Continuity of Sums and Products of Real-Valued Functions on a Metric Space (i) and Continuity of Sums and Products of Real-Valued Functions on a Metric Space (v)), and the hypothesis reads $u(t)\le w(t)$ for every $t\in[a,b]$.

Step 2: the majorant satisfies the same inequality. Since $w$ is continuous, $w\in\mathcal{L}^{1}([a,b])$ and $\int_{[a,t]}w\,d\lambda=\int_{a}^{t}w$, the oriented Riemann integral, for every $t\in[a,b]$ (Indefinite Lebesgue Integrals on a Closed Interval (xi)). As $u\le w$ on $[a,t]$ and $\beta\ge0$, monotonicity of the integral (Indefinite Lebesgue Integrals on a Closed Interval (iv)) gives

$$w(t)=\alpha+\beta\int_{[a,t]}u\,d\lambda\le\alpha+\beta\int_{[a,t]}w\,d\lambda=\alpha+\beta\int_{a}^{t}w\qquad(t\in[a,b]).$$

Step 3: conclusion. By Gronwall's lemma for continuous functions (Gronwall's Lemma (Integral Form) (i)), applied to $w$, $w(t)\le\alpha\exp(\beta(t-a))$ for every $t\in[a,b]$. Hence

$$u(t)\le w(t)\le\alpha\exp\bigl(\beta(t-a)\bigr)\qquad(t\in[a,b]).$$