[proofplan] We adapt the integrating-factor argument used for the homogeneous [Gronwall Inequality](/theorems/3377). Define $\mu(t) := \exp(-\int_0^t \beta\,d\mathcal{L}^1)$; since $\beta$ is integrable, $\mu$ is absolutely continuous, strictly positive, and satisfies $\mu' = -\beta\mu$ almost everywhere. Multiplying the differential inequality by $\mu$ converts the left-hand side into the [derivative](/page/Derivative) of the absolutely continuous product $\phi\mu$, with upper bound $\mu\eta$ rather than $0$. Applying the fundamental theorem of calculus for absolutely continuous functions and dividing by $\mu(t) > 0$ then produces the closed-form bound, with the inhomogeneity contributing the integral term $\int_0^t \eta(s)\,\mu(s)/\mu(t)\,d\mathcal{L}^1(s)$ that distinguishes this result from the homogeneous case. [/proofplan]

[step:Build the integrating factor $\mu(t) = \exp(-\int_0^t \beta\,d\mathcal{L}^1)$ as an absolutely continuous, strictly positive [function](/page/Function)] Define \begin{align*} B: [0,T] &\to \mathbb{R} \\ t &\mapsto \int_0^t \beta(s)\,d\mathcal{L}^1(s), \end{align*} and \begin{align*} \mu: [0,T] &\to (0,\infty) \\ t &\mapsto \exp(-B(t)). \end{align*} Since $\beta: [0,T] \to \mathbb{R}$ is [integrable](/page/Integral), the [Fundamental Theorem of Calculus for Lebesgue Integrals](/theorems/???) applies to $B$, so $B$ is absolutely [continuous](/page/Continuity) on $[0,T]$ with $B'(t) = \beta(t)$ for $\mathcal{L}^1$-a.e.\ $t \in [0,T]$ and $B(0) = 0$. The exponential map $\exp: \mathbb{R} \to (0,\infty)$ is $C^1$, hence locally Lipschitz, so by the [Chain Rule for Absolutely Continuous Functions](/theorems/???) the composition $\mu = \exp \circ (-B)$ is absolutely continuous on $[0,T]$ with \begin{align*} \mu'(t) &= -B'(t)\exp(-B(t)) = -\beta(t)\,\mu(t) \qquad \text{for $\mathcal{L}^1$-a.e.\ } t \in [0,T]. \end{align*} Finally $\mu(0) = \exp(0) = 1$ and $\mu(t) > 0$ for every $t \in [0,T]$.[/step]

[guided]The key idea is to construct an auxiliary function whose product with $\phi$ converts the differential inequality $\phi' \le \beta\phi + \eta$ into one whose left-hand side is a pure derivative. This is the standard integrating-factor technique from linear first-order ODEs. We want $\mu$ such that $(\phi\mu)' = \mu(\phi' - \beta\phi)$, which forces $\mu' = -\beta\mu$. Solving formally gives $\mu(t) = \exp(-\int_0^t \beta\,d\mathcal{L}^1)$. We make this rigorous. Define \begin{align*} B: [0,T] &\to \mathbb{R} \\ t &\mapsto \int_0^t \beta(s)\,d\mathcal{L}^1(s), \end{align*} and \begin{align*} \mu: [0,T] &\to (0,\infty) \\ t &\mapsto \exp(-B(t)). \end{align*} **Showing $B$ is absolutely continuous with $B' = \beta$ a.e.** We invoke the [Fundamental Theorem of Calculus for Lebesgue Integrals](/theorems/???), which states: if $f \in L^1([0,T])$, then $F(t) := \int_0^t f\,d\mathcal{L}^1$ is absolutely continuous on $[0,T]$ with $F'(t) = f(t)$ for $\mathcal{L}^1$-a.e.\ $t$. The hypothesis is $\beta \in L^1([0,T])$, which is exactly the integrability assumption on $\beta$ in the theorem statement. Therefore $B$ is absolutely continuous with $B'(t) = \beta(t)$ a.e., and $B(0) = \int_0^0 \beta\,d\mathcal{L}^1 = 0$. **Showing $\mu$ is absolutely continuous with $\mu' = -\beta\mu$ a.e.** We compose $B$ with $\exp$. The relevant tool is the [Chain Rule for Absolutely Continuous Functions](/theorems/???): if $g: [a,b] \to [c,d]$ is absolutely continuous and $h: [c,d] \to \mathbb{R}$ is locally Lipschitz (equivalently, $C^1$ with bounded derivative on every compact subinterval), then $h \circ g$ is absolutely continuous with $(h \circ g)'(t) = h'(g(t))\,g'(t)$ a.e. We verify: $-B$ is absolutely continuous (negation preserves absolute continuity), its range is the compact interval $[-\|B\|_\infty, \|B\|_\infty]$ (note $B$ is continuous on the compact interval $[0,T]$, hence bounded), and $\exp$ is $C^1$ with derivative $\exp$, locally bounded. The chain rule gives \begin{align*} \mu'(t) &= \exp(-B(t)) \cdot (-B'(t)) = -\beta(t)\,\mu(t) \qquad \text{for $\mathcal{L}^1$-a.e.\ } t \in [0,T]. \end{align*} **Positivity and boundary value.** The exponential is strictly positive, so $\mu(t) > 0$ for every $t \in [0,T]$. At $t = 0$, $\mu(0) = \exp(-B(0)) = \exp(0) = 1$. Positivity is essential because Step 4 divides by $\mu(t)$; the boundary value $\mu(0) = 1$ will be used in Step 3 to extract the constant $\alpha$.[/guided]

[step:Differentiate the product $\phi\mu$ and bound it using the differential inequality]Both $\phi$ and $\mu$ are absolutely continuous on $[0,T]$ ($\phi$ by hypothesis, $\mu$ by Step 1). By the [Product Rule for Absolutely Continuous Functions](/theorems/???), the product $\phi\mu: [0,T] \to \mathbb{R}$ is absolutely continuous on $[0,T]$ with \begin{align*} (\phi\mu)'(t) &= \phi'(t)\,\mu(t) + \phi(t)\,\mu'(t) \qquad \text{for $\mathcal{L}^1$-a.e.\ } t \in [0,T]. \end{align*} Substituting $\mu'(t) = -\beta(t)\,\mu(t)$ from Step 1, \begin{align*} (\phi\mu)'(t) &= \mu(t)\bigl[\phi'(t) - \beta(t)\,\phi(t)\bigr] \qquad \text{for $\mathcal{L}^1$-a.e.\ } t \in [0,T]. \end{align*} By hypothesis $\phi'(t) - \beta(t)\,\phi(t) \le \eta(t)$ for $\mathcal{L}^1$-a.e.\ $t$, and $\mu(t) > 0$ for every $t$, so multiplying preserves the inequality: \begin{align*} (\phi\mu)'(t) &\le \mu(t)\,\eta(t) \qquad \text{for $\mathcal{L}^1$-a.e.\ } t \in [0,T]. \end{align*}[/step]

[guided]With the integrating factor $\mu$ in hand, we compute $(\phi\mu)'$ and exploit the differential inequality. Why a product? Because the structure $\phi' - \beta\phi$ on the left of the inequality is precisely what arises when one differentiates $\phi\mu$ and uses $\mu' = -\beta\mu$. **Applying the product rule.** Both $\phi$ and $\mu$ are absolutely continuous on $[0,T]$: $\phi$ by the theorem's hypothesis, and $\mu$ by Step 1. We apply the [Product Rule for Absolutely Continuous Functions](/theorems/???): if $f, g: [a,b] \to \mathbb{R}$ are both absolutely continuous, then $fg$ is absolutely continuous and \begin{align*} (fg)'(t) &= f'(t)\,g(t) + f(t)\,g'(t) \qquad \text{for $\mathcal{L}^1$-a.e.\ } t \in [a,b]. \end{align*} Both hypotheses are verified, so the conclusion applies to $\phi\mu$: \begin{align*} (\phi\mu)'(t) &= \phi'(t)\,\mu(t) + \phi(t)\,\mu'(t) \qquad \text{for $\mathcal{L}^1$-a.e.\ } t \in [0,T]. \end{align*} **Substituting $\mu' = -\beta\mu$.** From Step 1 we have $\mu'(t) = -\beta(t)\,\mu(t)$ a.e. The set where both identities hold simultaneously is still of full measure (intersection of two full-measure sets). Substituting and factoring out $\mu(t)$: \begin{align*} (\phi\mu)'(t) &= \phi'(t)\,\mu(t) - \beta(t)\,\phi(t)\,\mu(t) = \mu(t)\bigl[\phi'(t) - \beta(t)\,\phi(t)\bigr] \end{align*} for $\mathcal{L}^1$-a.e.\ $t \in [0,T]$. This is exactly the structure we engineered the integrating factor to produce. **Using the differential inequality.** The hypothesis $\phi'(t) \le \beta(t)\,\phi(t) + \eta(t)$ a.e.\ rearranges to \begin{align*} \phi'(t) - \beta(t)\,\phi(t) &\le \eta(t) \qquad \text{for $\mathcal{L}^1$-a.e.\ } t \in [0,T]. \end{align*} We multiply both sides by $\mu(t)$. Since $\mu(t) > 0$ for every $t \in [0,T]$ (Step 1), multiplication by $\mu(t)$ preserves the direction of the inequality at each $t$. Multiplying: \begin{align*} (\phi\mu)'(t) &= \mu(t)\bigl[\phi'(t) - \beta(t)\,\phi(t)\bigr] \le \mu(t)\,\eta(t) \end{align*} for $\mathcal{L}^1$-a.e.\ $t \in [0,T]$. Note that no sign assumption on $\eta$ is needed — the bound holds whether $\eta$ is positive, negative, or signed.[/guided]

[step:Integrate the differential inequality from $0$ to $t$ using the fundamental theorem of calculus]Since $\phi\mu$ is absolutely continuous on $[0,T]$ (Step 2), by the [Fundamental Theorem of Calculus for Absolutely Continuous Functions](/theorems/???), for every $t \in [0,T]$, \begin{align*} \phi(t)\,\mu(t) - \phi(0)\,\mu(0) &= \int_0^t (\phi\mu)'(s)\,d\mathcal{L}^1(s). \end{align*} The pointwise inequality $(\phi\mu)'(s) \le \mu(s)\,\eta(s)$ holds for $\mathcal{L}^1$-a.e.\ $s \in [0,t]$ (Step 2). Both $(\phi\mu)'$ and $\mu\eta$ are integrable on $[0,t]$: $(\phi\mu)'$ because $\phi\mu$ is absolutely continuous, and $\mu\eta$ because $\mu$ is continuous on the compact interval $[0,T]$ (hence bounded) and $\eta \in L^1([0,T])$ by hypothesis. By [monotonicity of the Lebesgue integral](/theorems/???), \begin{align*} \int_0^t (\phi\mu)'(s)\,d\mathcal{L}^1(s) &\le \int_0^t \mu(s)\,\eta(s)\,d\mathcal{L}^1(s). \end{align*} Combining, and using $\mu(0) = 1$ and $\phi(0) \le \alpha$, \begin{align*} \phi(t)\,\mu(t) &= \phi(0)\,\mu(0) + \int_0^t (\phi\mu)'(s)\,d\mathcal{L}^1(s) \le \alpha + \int_0^t \mu(s)\,\eta(s)\,d\mathcal{L}^1(s). \end{align*}[/step]

[guided]We now convert the differential inequality $(\phi\mu)'(s) \le \mu(s)\,\eta(s)$ into an integral inequality by integrating from $0$ to $t$. **Recovering $\phi(t)\mu(t) - \phi(0)\mu(0)$ via FTC.** We invoke the [Fundamental Theorem of Calculus for Absolutely Continuous Functions](/theorems/???): if $F: [a,b] \to \mathbb{R}$ is absolutely continuous, then $F'$ exists $\mathcal{L}^1$-a.e., $F' \in L^1([a,b])$, and \begin{align*} F(t) - F(a) &= \int_a^t F'(s)\,d\mathcal{L}^1(s) \qquad \text{for every } t \in [a,b]. \end{align*} We apply this to $F = \phi\mu$ on $[0,t]$. The hypothesis — absolute continuity of $\phi\mu$ — was established in Step 2. Therefore \begin{align*} \phi(t)\,\mu(t) - \phi(0)\,\mu(0) &= \int_0^t (\phi\mu)'(s)\,d\mathcal{L}^1(s) \qquad \text{for every } t \in [0,T]. \end{align*} **Integrating the a.e.\ inequality.** From Step 2, $(\phi\mu)'(s) \le \mu(s)\,\eta(s)$ for $\mathcal{L}^1$-a.e.\ $s \in [0,T]$. We integrate this inequality over $[0,t]$ using [monotonicity of the Lebesgue integral](/theorems/???): if $f, g \in L^1([0,t])$ with $f \le g$ a.e., then $\int_0^t f\,d\mathcal{L}^1 \le \int_0^t g\,d\mathcal{L}^1$. Verifying integrability: - $(\phi\mu)' \in L^1([0,t])$ as a consequence of the FTC applied above. - $\mu\eta \in L^1([0,t])$: $\mu$ is continuous on the compact interval $[0,T]$ (continuity follows from absolute continuity), so $\mu$ is bounded, say $|\mu(s)| \le M$ for $s \in [0,T]$. Then $|\mu(s)\,\eta(s)| \le M\,|\eta(s)|$, and $\eta \in L^1([0,T])$ by hypothesis, so $\mu\eta \in L^1([0,T]) \subseteq L^1([0,t])$. Both integrability conditions hold, so monotonicity gives \begin{align*} \int_0^t (\phi\mu)'(s)\,d\mathcal{L}^1(s) &\le \int_0^t \mu(s)\,\eta(s)\,d\mathcal{L}^1(s). \end{align*} **Applying the boundary conditions.** From Step 1, $\mu(0) = 1$. From the theorem's hypothesis, $\phi(0) \le \alpha$. Therefore $\phi(0)\,\mu(0) = \phi(0) \le \alpha$. Chaining the equality and inequality: \begin{align*} \phi(t)\,\mu(t) &= \phi(0)\,\mu(0) + \int_0^t (\phi\mu)'(s)\,d\mathcal{L}^1(s) \\ &\le \alpha + \int_0^t \mu(s)\,\eta(s)\,d\mathcal{L}^1(s) \qquad \text{for every } t \in [0,T]. \end{align*}[/guided]

[step:Divide by $\mu(t) > 0$ and rewrite $\mu(s)/\mu(t)$ in exponential form]Fix $t \in [0,T]$. Since $\mu(t) > 0$ (Step 1), dividing the inequality from Step 3 by $\mu(t)$ preserves its direction: \begin{align*} \phi(t) &\le \frac{\alpha}{\mu(t)} + \int_0^t \frac{\mu(s)}{\mu(t)}\,\eta(s)\,d\mathcal{L}^1(s). \end{align*} (Note: $\mu(t)$ is a positive constant relative to the integration variable $s$, so it can be moved inside the integral as $1/\mu(t)$.) We now rewrite the two ratios in exponential form. By definition $\mu(t) = \exp(-B(t))$, so \begin{align*} \frac{1}{\mu(t)} &= \exp(B(t)) = \exp\!\Bigl(\int_0^t \beta(r)\,d\mathcal{L}^1(r)\Bigr). \end{align*} Similarly, using the homomorphism property $\exp(a)\exp(b) = \exp(a+b)$, \begin{align*} \frac{\mu(s)}{\mu(t)} &= \frac{\exp(-B(s))}{\exp(-B(t))} = \exp(-B(s) + B(t)) = \exp(B(t) - B(s)) \\ &= \exp\!\Bigl(\int_0^t \beta(r)\,d\mathcal{L}^1(r) - \int_0^s \beta(r)\,d\mathcal{L}^1(r)\Bigr) = \exp\!\Bigl(\int_s^t \beta(r)\,d\mathcal{L}^1(r)\Bigr), \end{align*} where the last equality uses additivity of the [Lebesgue integral](/page/Lebesgue Integral) over the disjoint decomposition $[0,t] = [0,s] \cup [s,t]$ (valid for every $s \in [0,t]$). Substituting both identities, \begin{align*} \phi(t) &\le \alpha\,\exp\!\Bigl(\int_0^t \beta(r)\,d\mathcal{L}^1(r)\Bigr) + \int_0^t \eta(s)\,\exp\!\Bigl(\int_s^t \beta(r)\,d\mathcal{L}^1(r)\Bigr)\,d\mathcal{L}^1(s) \end{align*} for every $t \in [0,T]$, which is the desired conclusion.[/step]

[guided]The work is essentially done; what remains is to translate the bound on $\phi(t)\mu(t)$ from Step 3 into a bound on $\phi(t)$, and to rewrite the integrating factor ratios in the explicit exponential form stated in the theorem. **Dividing by $\mu(t) > 0$.** Fix $t \in [0,T]$. By Step 1, $\mu(t) > 0$, so $1/\mu(t) > 0$ is a positive constant (with respect to the integration variable $s$, which ranges over $[0,t]$). Multiplying the inequality from Step 3 by $1/\mu(t)$ preserves its direction: \begin{align*} \phi(t) &= \frac{\phi(t)\,\mu(t)}{\mu(t)} \le \frac{\alpha + \int_0^t \mu(s)\,\eta(s)\,d\mathcal{L}^1(s)}{\mu(t)} = \frac{\alpha}{\mu(t)} + \int_0^t \frac{\mu(s)}{\mu(t)}\,\eta(s)\,d\mathcal{L}^1(s). \end{align*} We pulled $1/\mu(t)$ inside the integral by linearity, which is valid because $\mu(t)$ is a constant relative to $s$ — it depends only on the fixed upper limit $t$. **Rewriting $1/\mu(t)$.** By definition $\mu(t) = \exp(-B(t))$, so \begin{align*} \frac{1}{\mu(t)} &= \frac{1}{\exp(-B(t))} = \exp(B(t)) = \exp\!\Bigl(\int_0^t \beta(r)\,d\mathcal{L}^1(r)\Bigr). \end{align*} **Rewriting $\mu(s)/\mu(t)$ — the key algebraic identity.** This is the heart of the closed form. The exponential is a homomorphism from $(\mathbb{R}, +)$ to $((0,\infty), \cdot)$, meaning $\exp(a + b) = \exp(a)\exp(b)$ for all $a, b \in \mathbb{R}$. Equivalently, $\exp(a)/\exp(b) = \exp(a - b)$. Applying this with $a = -B(s)$ and $b = -B(t)$: \begin{align*} \frac{\mu(s)}{\mu(t)} &= \frac{\exp(-B(s))}{\exp(-B(t))} = \exp(-B(s) - (-B(t))) = \exp(B(t) - B(s)). \end{align*} Now we compute $B(t) - B(s)$. By definition of $B$, \begin{align*} B(t) - B(s) &= \int_0^t \beta(r)\,d\mathcal{L}^1(r) - \int_0^s \beta(r)\,d\mathcal{L}^1(r). \end{align*} For $0 \le s \le t \le T$, additivity of the Lebesgue integral over the disjoint decomposition $[0,t] = [0,s] \sqcup [s,t]$ (up to the single point $\{s\}$, which has Lebesgue measure zero) gives $\int_0^t \beta\,d\mathcal{L}^1 = \int_0^s \beta\,d\mathcal{L}^1 + \int_s^t \beta\,d\mathcal{L}^1$, so \begin{align*} B(t) - B(s) &= \int_s^t \beta(r)\,d\mathcal{L}^1(r). \end{align*} Therefore \begin{align*} \frac{\mu(s)}{\mu(t)} &= \exp\!\Bigl(\int_s^t \beta(r)\,d\mathcal{L}^1(r)\Bigr). \end{align*} **Final substitution.** Plugging both identities back into the inequality: \begin{align*} \phi(t) &\le \alpha\,\exp\!\Bigl(\int_0^t \beta(r)\,d\mathcal{L}^1(r)\Bigr) + \int_0^t \eta(s)\,\exp\!\Bigl(\int_s^t \beta(r)\,d\mathcal{L}^1(r)\Bigr)\,d\mathcal{L}^1(s). \end{align*} This holds for every $t \in [0,T]$ (we fixed $t$ arbitrarily at the start), which is exactly the conclusion of the theorem. Note that $\eta$ remains inside the integral with its original sign — at no point in the argument did we use $\eta \ge 0$, so the inhomogeneous term in the conclusion correctly carries whatever sign $\eta$ has.[/guided]