[proofplan] The proof uses the integrating-factor technique from the theory of first-order linear ODEs, adapted to an inequality setting. We define an auxiliary [function](/page/Function) $\Gamma$ that dominates $\phi$ and satisfies a differential inequality $\Gamma' \le \beta\,\Gamma$ almost everywhere. Introducing a non-negative defect $q := \beta\,\Gamma - \Gamma'$ and multiplying by the integrating factor $\mu(t) = \exp\bigl(-\int_0^t \beta\,d\mathcal{L}^1\bigr)$, the product rule converts the left-hand side into the exact derivative $\frac{d}{dt}[\Gamma\,\mu]$, which is shown to be non-positive. Since an absolutely continuous function with non-positive derivative almost everywhere is non-increasing, we conclude $\Gamma(t)\,\mu(t) \le \Gamma(0)\,\mu(0) = \alpha$, and the bound on $\phi$ follows from $\phi \le \Gamma$. [/proofplan] [step:Define the auxiliary function $\Gamma$ dominating $\phi$] Define \begin{align*} \Gamma: [0,T] &\to [0,\infty) \\ t &\mapsto \alpha + \int_0^t \beta(s)\,\phi(s)\,d\mathcal{L}^1(s). \end{align*} Then $\Gamma(0) = \alpha$. Since $\beta$ is [integrable](/page/Integral) and $\phi$ is continuous on the [compact](/page/Compact%20Space) interval $[0,T]$, the product $\beta\,\phi$ is integrable, so $\Gamma$ is absolutely [continuous](/page/Continuity). The hypothesis $\phi(t) \le \alpha + \int_0^t \beta(s)\,\phi(s)\,d\mathcal{L}^1(s)$ gives \begin{align*} \phi(t) &\le \Gamma(t) \quad \text{for all } t \in [0,T]. \end{align*} [guided] We want to convert the integral inequality into a differential one so that classical ODE techniques apply. The natural move is to package the right-hand side of the hypothesis into a single function. Define \begin{align*} \Gamma: [0,T] &\to [0,\infty) \\ t &\mapsto \alpha + \int_0^t \beta(s)\,\phi(s)\,d\mathcal{L}^1(s). \end{align*} We check that $\Gamma$ is well-defined and has the regularity we need. Since $\phi: [0,T] \to [0,\infty)$ is continuous on the compact interval $[0,T]$, it is bounded, so the product $\beta\,\phi$ is integrable (the product of an integrable function and a bounded [measurable](/page/Measurable%20Functions) function is integrable). Therefore $\Gamma$ is the sum of a constant $\alpha$ and an indefinite [Lebesgue integral](/page/Lebesgue%20Integral) of an integrable function, which makes $\Gamma$ absolutely continuous on $[0,T]$. Evaluating at $t = 0$: the integral vanishes, so $\Gamma(0) = \alpha$. The hypothesis of the theorem states $\phi(t) \le \alpha + \int_0^t \beta(s)\,\phi(s)\,d\mathcal{L}^1(s)$ for all $t \in [0,T]$. The right-hand side is exactly $\Gamma(t)$, so \begin{align*} \phi(t) &\le \Gamma(t) \quad \text{for all } t \in [0,T]. \end{align*} Why define $\Gamma$? The function $\Gamma$ captures the "worst case" allowed by the integral inequality. By working with $\Gamma$ instead of $\phi$, we obtain a closed differential inequality involving $\Gamma$ alone, which is amenable to the integrating-factor method. [/guided] [/step] [step:Derive the differential inequality $\Gamma' \le \beta\,\Gamma$] Since $\Gamma$ is absolutely continuous, it is [differentiable](/page/Derivative) $\mathcal{L}^1$-a.e. on $[0,T]$, and by the Lebesgue differentiation theorem its derivative equals the integrand: \begin{align*} \Gamma'(t) &= \beta(t)\,\phi(t) \end{align*} for $\mathcal{L}^1$-a.e. $t \in [0,T]$. Since $\phi(t) \le \Gamma(t)$ from the previous step and $\beta(t) \ge 0$, we obtain \begin{align*} \Gamma'(t) &= \beta(t)\,\phi(t) \le \beta(t)\,\Gamma(t) \end{align*} for $\mathcal{L}^1$-a.e. $t \in [0,T]$. [guided] With $\Gamma$ in hand, we differentiate to convert the integral relation into a pointwise one. Since $\Gamma$ is the sum of a constant and the indefinite integral $t \mapsto \int_0^t \beta(s)\,\phi(s)\,d\mathcal{L}^1(s)$, and since $\beta\,\phi$ is integrable, the fundamental theorem of calculus for Lebesgue integrals (the fact that the derivative of an indefinite integral equals the integrand $\mathcal{L}^1$-a.e.) gives \begin{align*} \Gamma'(t) &= \beta(t)\,\phi(t) \end{align*} for $\mathcal{L}^1$-a.e. $t \in [0,T]$. Now we use the bound $\phi(t) \le \Gamma(t)$ established in the previous step. Since $\beta(t) \ge 0$ (by hypothesis, $\beta$ takes values in $[0,\infty)$), multiplying the inequality $\phi(t) \le \Gamma(t)$ by $\beta(t)$ preserves the direction: \begin{align*} \Gamma'(t) &= \beta(t)\,\phi(t) \le \beta(t)\,\Gamma(t) \end{align*} for $\mathcal{L}^1$-a.e. $t \in [0,T]$. This is a differential inequality of the same form as the ODE $y' = \beta\,y$, but with "$\le$" instead of "$=$". The next step applies the integrating-factor technique to extract a bound from this inequality. [/guided] [/step] [step:Multiply by the integrating factor to show $\frac{d}{dt}[\Gamma\,\mu] \le 0$] Define the non-negative defect \begin{align*} q: [0,T] &\to [0,\infty) \\ t &\mapsto \beta(t)\,\Gamma(t) - \Gamma'(t). \end{align*} By the previous step, $q(t) \ge 0$ for $\mathcal{L}^1$-a.e. $t \in [0,T]$. Rearranging: \begin{align*} \Gamma'(t) - \beta(t)\,\Gamma(t) &= -q(t). \end{align*} Define the [integrating factor](/theorems/890) \begin{align*} \mu: [0,T] &\to (0,\infty) \\ t &\mapsto \exp\!\Bigl(-\int_0^t \beta(s)\,d\mathcal{L}^1(s)\Bigr). \end{align*} Since $\beta \ge 0$ and $\beta$ is integrable, the exponent $-\int_0^t \beta\,d\mathcal{L}^1$ is non-positive, so $\mu(t) \in (0,1]$ for all $t$. The function $\mu$ is absolutely continuous with \begin{align*} \mu'(t) &= -\beta(t)\,\mu(t) \end{align*} for $\mathcal{L}^1$-a.e. $t \in [0,T]$ (by the chain rule for absolutely continuous functions). Multiplying the rearranged equation by $\mu(t) > 0$: \begin{align*} \mu(t)\,\Gamma'(t) + \mu'(t)\,\Gamma(t) &= \mu(t)\,\Gamma'(t) - \beta(t)\,\mu(t)\,\Gamma(t) = -q(t)\,\mu(t). \end{align*} By the product rule for absolutely continuous functions, the left-hand side equals $\frac{d}{dt}[\Gamma(t)\,\mu(t)]$. Since $q(t) \ge 0$ and $\mu(t) > 0$: \begin{align*} \frac{d}{dt}\bigl[\Gamma(t)\,\mu(t)\bigr] &= -q(t)\,\mu(t) \le 0 \end{align*} for $\mathcal{L}^1$-a.e. $t \in [0,T]$. [guided] The differential inequality $\Gamma' \le \beta\,\Gamma$ looks like a linear first-order ODE, and the standard technique for such ODEs is the integrating factor. The idea is to multiply by a function chosen so that the left-hand side becomes an exact derivative. **Introducing the defect.** Rather than working directly with the inequality, we introduce the "slack" as a named quantity. Define \begin{align*} q: [0,T] &\to [0,\infty) \\ t &\mapsto \beta(t)\,\Gamma(t) - \Gamma'(t). \end{align*} The differential inequality $\Gamma'(t) \le \beta(t)\,\Gamma(t)$ from the previous step says exactly that $q(t) \ge 0$ for $\mathcal{L}^1$-a.e. $t \in [0,T]$. Rearranging gives an equation (not just an inequality): \begin{align*} \Gamma'(t) - \beta(t)\,\Gamma(t) &= -q(t). \end{align*} **Choosing the integrating factor.** For the ODE $y' - \beta\,y = 0$, the integrating factor is $\mu(t) = \exp\bigl(-\int_0^t \beta(s)\,d\mathcal{L}^1(s)\bigr)$, because $\mu$ is designed so that $\mu' = -\beta\,\mu$. Define \begin{align*} \mu: [0,T] &\to (0,\infty) \\ t &\mapsto \exp\!\Bigl(-\int_0^t \beta(s)\,d\mathcal{L}^1(s)\Bigr). \end{align*} We verify the required properties. Since $\beta \ge 0$ is integrable, the map $t \mapsto \int_0^t \beta(s)\,d\mathcal{L}^1(s)$ is absolutely continuous and non-decreasing, so $\mu$ is the composition of an absolutely continuous function with the smooth function $\exp$, hence absolutely continuous. The chain rule for absolutely continuous functions gives \begin{align*} \mu'(t) &= -\beta(t)\,\mu(t) \end{align*} for $\mathcal{L}^1$-a.e. $t \in [0,T]$. Note that $\mu(t) > 0$ for all $t$ (the exponential is always positive), and $\mu(0) = \exp(0) = 1$. **Multiplying and applying the product rule.** We multiply the equation $\Gamma'(t) - \beta(t)\,\Gamma(t) = -q(t)$ by $\mu(t) > 0$: \begin{align*} \mu(t)\,\Gamma'(t) - \beta(t)\,\mu(t)\,\Gamma(t) &= -q(t)\,\mu(t). \end{align*} Since $\mu'(t) = -\beta(t)\,\mu(t)$, the left-hand side becomes $\mu(t)\,\Gamma'(t) + \mu'(t)\,\Gamma(t)$. Both $\Gamma$ and $\mu$ are absolutely continuous, so the product rule for absolutely continuous functions applies, giving \begin{align*} \frac{d}{dt}\bigl[\Gamma(t)\,\mu(t)\bigr] &= \mu(t)\,\Gamma'(t) + \mu'(t)\,\Gamma(t) = -q(t)\,\mu(t). \end{align*} Since $q(t) \ge 0$ and $\mu(t) > 0$, the right-hand side is non-positive: \begin{align*} \frac{d}{dt}\bigl[\Gamma(t)\,\mu(t)\bigr] &\le 0 \end{align*} for $\mathcal{L}^1$-a.e. $t \in [0,T]$. Why did we introduce the defect $q$ instead of just "multiplying the inequality by $\mu$"? Working with the equation $\Gamma' - \beta\,\Gamma = -q$ makes each algebraic step an equality, so the only place the inequality enters is in the final observation $q \ge 0$. This cleanly separates the algebraic manipulation (exact) from the order-theoretic content (the sign of $q$). [/guided] [/step] [step:Integrate and conclude $\phi(t) \le \alpha\,\exp\!\bigl(\int_0^t \beta\,d\mathcal{L}^1\bigr)$] The function $t \mapsto \Gamma(t)\,\mu(t)$ is absolutely continuous (as a product of absolutely continuous functions) with non-positive derivative $\mathcal{L}^1$-a.e., so it is non-increasing on $[0,T]$. Therefore, for all $t \in [0,T]$: \begin{align*} \Gamma(t)\,\mu(t) &\le \Gamma(0)\,\mu(0) = \alpha \cdot 1 = \alpha. \end{align*} Dividing both sides by $\mu(t) > 0$: \begin{align*} \Gamma(t) &\le \frac{\alpha}{\mu(t)} = \alpha\,\exp\!\Bigl(\int_0^t \beta(s)\,d\mathcal{L}^1(s)\Bigr). \end{align*} Since $\phi(t) \le \Gamma(t)$ for all $t \in [0,T]$ (from the first step), we conclude \begin{align*} \phi(t) &\le \alpha\,\exp\!\Bigl(\int_0^t \beta(s)\,d\mathcal{L}^1(s)\Bigr), \quad \forall\, t \in [0,T]. \end{align*} [guided] We have shown that the absolutely continuous function $t \mapsto \Gamma(t)\,\mu(t)$ has a non-positive derivative $\mathcal{L}^1$-a.e. on $[0,T]$. An absolutely continuous function whose derivative is non-positive almost everywhere is non-increasing (this follows from the fundamental theorem of calculus for absolutely continuous functions: $\Gamma(t)\,\mu(t) - \Gamma(0)\,\mu(0) = \int_0^t \frac{d}{ds}[\Gamma(s)\,\mu(s)]\,d\mathcal{L}^1(s) \le 0$). Therefore, for all $t \in [0,T]$: \begin{align*} \Gamma(t)\,\mu(t) &\le \Gamma(0)\,\mu(0). \end{align*} We evaluate the right-hand side: $\Gamma(0) = \alpha$ (computed in the first step) and $\mu(0) = \exp(0) = 1$, so $\Gamma(0)\,\mu(0) = \alpha$. Dividing both sides by $\mu(t) > 0$ (the exponential is strictly positive) and using $1/\mu(t) = \exp\bigl(\int_0^t \beta(s)\,d\mathcal{L}^1(s)\bigr)$: \begin{align*} \Gamma(t) &\le \alpha\,\exp\!\Bigl(\int_0^t \beta(s)\,d\mathcal{L}^1(s)\Bigr). \end{align*} Finally, chaining with $\phi(t) \le \Gamma(t)$ (established in the first step): \begin{align*} \phi(t) &\le \alpha\,\exp\!\Bigl(\int_0^t \beta(s)\,d\mathcal{L}^1(s)\Bigr), \quad \forall\, t \in [0,T]. \end{align*} This is the desired conclusion. [/guided] [/step]