[proofplan] The argument is the classical integrating-factor reduction. 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*} so that $\mu' = -\beta\mu$ holds $\mathcal{L}^1$-a.e. Multiplying $\phi$ by $\mu$ converts the differential inequality $\phi' \le \beta\,\phi$ into the assertion that $(\phi\mu)' \le 0$ $\mathcal{L}^1$-a.e., via the product rule for absolutely continuous [functions](/page/Function). The fundamental theorem of calculus for absolutely continuous functions then forces $\phi\mu$ to be non-increasing, so $\phi(t)\mu(t) \le \phi(0)\mu(0) \le \alpha$, and dividing by the positive quantity $\mu(t)$ produces the claimed exponential bound. [/proofplan] [step:Introduce the integrating factor $\mu(t) = \exp(-\int_0^t \beta\,d\mathcal{L}^1)$ and compute $\mu' = -\beta\mu$ a.e.] 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 $\beta \in L^1([0,T])$, the function $B$ is absolutely continuous on $[0,T]$ by the [Fundamental Theorem of Calculus for Lebesgue Integrals](/theorems/???), 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 $C^1$ (in particular locally Lipschitz), so by the [Chain Rule for Absolutely Continuous Functions](/theorems/???) applied to $\mu = \exp \circ (-B)$, $\mu$ is absolutely continuous on $[0,T]$ and \begin{align*} \mu'(t) &= -B'(t)\,\exp(-B(t)) = -\beta(t)\,\mu(t) \end{align*} for $\mathcal{L}^1$-a.e.\ $t \in [0,T]$. Moreover $\mu(0) = \exp(0) = 1$ and $\mu(t) = \exp(-B(t)) > 0$ for every $t \in [0,T]$ because the exponential is strictly positive. [guided] We want a quantity that absorbs the $\beta\phi$ term in the inequality $\phi' \le \beta\phi$ and converts it into a monotonicity statement. The classical recipe is an **integrating factor**: a strictly positive multiplier $\mu(t)$ whose [derivative](/page/Derivative) is $-\beta(t)\,\mu(t)$, so that the product rule produces a cancellation. The unique such $\mu$ (up to scaling) with $\mu(0) = 1$ is the exponential of the antiderivative $-\int_0^\cdot \beta\,d\mathcal{L}^1$. We build it carefully and verify each hypothesis needed to differentiate it. First, define the antiderivative \begin{align*} B: [0,T] &\to \mathbb{R} \\ t &\mapsto \int_0^t \beta(s)\,d\mathcal{L}^1(s). \end{align*} The codomain is $\mathbb{R}$ because $\beta$ may take any sign — note that the argument below never uses the sign of $\beta$, only its integrability. To apply the [Fundamental Theorem of Calculus for Lebesgue Integrals](/theorems/???) (which says: if $f \in L^1([0,T])$, then $F(t) := \int_0^t f\,d\mathcal{L}^1$ is absolutely continuous with $F' = f$ $\mathcal{L}^1$-a.e.), we must check $\beta \in L^1([0,T])$. This is exactly the [integrable](/page/Integral) hypothesis on $\beta$. The theorem therefore gives: - $B$ is absolutely continuous on $[0,T]$, - $B'(t) = \beta(t)$ for $\mathcal{L}^1$-a.e.\ $t \in [0,T]$, - $B(0) = 0$ (the integral over $[0,0]$ vanishes). Now define \begin{align*} \mu: [0,T] &\to (0,\infty) \\ t &\mapsto \exp(-B(t)). \end{align*} Why is the codomain $(0,\infty)$? Because $\exp: \mathbb{R} \to (0,\infty)$ is strictly positive, so $\mu(t) = \exp(-B(t)) > 0$ for every $t \in [0,T]$, regardless of the sign of $B(t)$. To compute $\mu'$, we want to apply the [Chain Rule for Absolutely Continuous Functions](/theorems/???): if $g: [0,T] \to \mathbb{R}$ is absolutely continuous and $F: \mathbb{R} \to \mathbb{R}$ is $C^1$ (or merely locally Lipschitz, which is enough), then $F \circ g$ is absolutely continuous with $(F \circ g)'(t) = F'(g(t))\,g'(t)$ for $\mathcal{L}^1$-a.e.\ $t$. Here $g = -B$ is absolutely continuous (negation preserves absolute continuity), and $F = \exp$ is $C^1$ on $\mathbb{R}$ with $F' = \exp$. Both hypotheses hold, so the chain rule applies and yields \begin{align*} \mu'(t) &= \exp(-B(t)) \cdot (-B'(t)) = -\beta(t)\,\mu(t) \end{align*} for $\mathcal{L}^1$-a.e.\ $t \in [0,T]$. Why insist on the AC chain rule rather than the classical $C^1$ chain rule? Because $B$ need not be differentiable everywhere — only $\mathcal{L}^1$-a.e. The AC chain rule is the correct tool whenever one of the factors is only differentiable a.e. but is still absolutely continuous. Finally, record two facts we will need later: $\mu(0) = \exp(0) = 1$, and $\mu(t) > 0$ for every $t \in [0,T]$ (strict positivity of the exponential). Strict positivity is what allows us to divide by $\mu(t)$ in the last step. [/guided] [/step] [step:Apply the product rule to show $(\phi\mu)' \le 0$ $\mathcal{L}^1$-a.e.] Since $\phi$ and $\mu$ are both absolutely continuous on $[0,T]$, their product $\phi\mu: [0,T] \to \mathbb{R}$ is absolutely continuous, and by the [Product Rule for Absolutely Continuous Functions](/theorems/???), \begin{align*} (\phi\mu)'(t) &= \phi'(t)\,\mu(t) + \phi(t)\,\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.\ $t \in [0,T]$, using $\mu'(t) = -\beta(t)\,\mu(t)$ from the previous step. The hypothesis $\phi'(t) \le \beta(t)\,\phi(t)$ for $\mathcal{L}^1$-a.e.\ $t$ gives $\phi'(t) - \beta(t)\,\phi(t) \le 0$ a.e., and $\mu(t) > 0$ everywhere, so \begin{align*} (\phi\mu)'(t) &\le 0 \end{align*} for $\mathcal{L}^1$-a.e.\ $t \in [0,T]$. [guided] We have built $\mu$ so that $\mu' = -\beta\mu$. Now we test whether the product $\phi\mu$ behaves the way we hoped — its derivative should pick up the inequality $\phi' \le \beta\phi$ as a sign condition. To differentiate $\phi\mu$, we invoke the [Product Rule for Absolutely Continuous Functions](/theorems/???): if $f, g: [0,T] \to \mathbb{R}$ are both absolutely continuous, then so is $fg$, and $(fg)' = f'g + fg'$ for $\mathcal{L}^1$-a.e.\ $t \in [0,T]$. We verify the two hypotheses: - $\phi$ is absolutely continuous on $[0,T]$ by hypothesis of the theorem, - $\mu$ is absolutely continuous on $[0,T]$ by the previous step. Both hold, so the AC product rule applies. Computing: \begin{align*} (\phi\mu)'(t) &= \phi'(t)\,\mu(t) + \phi(t)\,\mu'(t). \end{align*} Substituting $\mu'(t) = -\beta(t)\,\mu(t)$: \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.\ $t$. This is the cancellation we were after: the bracket is exactly the quantity that the hypothesis bounds. By the differential inequality, $\phi'(t) - \beta(t)\,\phi(t) \le 0$ for $\mathcal{L}^1$-a.e.\ $t$. The multiplier $\mu(t)$ is strictly positive at every point of $[0,T]$ (see previous step), so multiplying a non-positive number by a positive one preserves the inequality: \begin{align*} (\phi\mu)'(t) &\le 0 \quad \text{for } \mathcal{L}^1\text{-a.e.\ } t \in [0,T]. \end{align*} A remark on what could fail. If $\phi$ were not assumed absolutely continuous, the product rule above would not apply — Lipschitz or even $C^1$ off a measure-zero set is not enough; absolute continuity is the precise condition that makes the FTC and product rule available for almost-everywhere derivatives. Note that the sign of $\beta$ is never used: the cancellation $\mu(\phi' - \beta\phi)$ and the positivity of $\mu$ (which comes from positivity of the exponential alone) do not depend on whether $\beta \ge 0$. [/guided] [/step] [step:Integrate the monotonicity and divide by $\mu(t) > 0$ to conclude] The function $\phi\mu: [0,T] \to \mathbb{R}$ is absolutely continuous with $(\phi\mu)' \le 0$ $\mathcal{L}^1$-a.e., so by the [Fundamental Theorem of Calculus for Absolutely Continuous Functions](/theorems/???) applied on $[0,t]$, \begin{align*} \phi(t)\,\mu(t) - \phi(0)\,\mu(0) &= \int_0^t (\phi\mu)'(s)\,d\mathcal{L}^1(s) \le 0 \end{align*} for every $t \in [0,T]$. Using $\mu(0) = 1$ and $\phi(0) \le \alpha$, \begin{align*} \phi(t)\,\mu(t) &\le \phi(0) \le \alpha. \end{align*} Since $\mu(t) > 0$ for every $t \in [0,T]$, division is valid and preserves the inequality: \begin{align*} \phi(t) &\le \frac{\alpha}{\mu(t)} = \alpha\,\exp(B(t)) = \alpha\,\exp\!\Bigl(\int_0^t \beta(s)\,d\mathcal{L}^1(s)\Bigr) \end{align*} for every $t \in [0,T]$, where the second equality is the definition $\mu(t) = \exp(-B(t))$ inverted, and the third is the definition of $B$. This is the claimed bound. [guided] We have a non-positive almost-everywhere derivative on an absolutely continuous function. The way to convert this pointwise-a.e. statement into a global comparison is the [Fundamental Theorem of Calculus for Absolutely Continuous Functions](/theorems/???): if $h: [0,T] \to \mathbb{R}$ is absolutely continuous, then for every $t \in [0,T]$, \begin{align*} h(t) - h(0) &= \int_0^t h'(s)\,d\mathcal{L}^1(s), \end{align*} where $h'$ is the $\mathcal{L}^1$-a.e.\ derivative. We apply this with $h = \phi\mu$. The hypothesis to verify is absolute continuity of $\phi\mu$, which we obtained in the previous step (product of two AC functions is AC). The FTC then yields, 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 integrand is non-positive $\mathcal{L}^1$-a.e., so the integral is non-positive (a basic property of the [Lebesgue integral](/page/Lebesgue%20Integral): $f \le 0$ a.e. $\implies \int f\,d\mathcal{L}^1 \le 0$). Therefore \begin{align*} \phi(t)\,\mu(t) &\le \phi(0)\,\mu(0). \end{align*} We now use the two boundary values established earlier: $\mu(0) = 1$ (from Step 1) and $\phi(0) \le \alpha$ (theorem hypothesis). Chaining, \begin{align*} \phi(t)\,\mu(t) &\le \phi(0)\cdot 1 = \phi(0) \le \alpha. \end{align*} The last move is to isolate $\phi(t)$. Because $\mu(t) > 0$ for every $t \in [0,T]$, we may divide both sides of $\phi(t)\,\mu(t) \le \alpha$ by $\mu(t)$ without flipping the inequality: \begin{align*} \phi(t) &\le \frac{\alpha}{\mu(t)}. \end{align*} This step is where strict positivity of $\mu$ is consumed; division by zero or by a quantity of indefinite sign would invalidate the conclusion. Why is this safe even when $\alpha < 0$? Because the algebra of dividing both sides of $\phi(t)\mu(t) \le \alpha$ by the positive quantity $\mu(t) > 0$ preserves the inequality regardless of the sign of $\alpha$; only positivity of $\mu(t)$ matters. Finally, $1/\mu(t) = 1/\exp(-B(t)) = \exp(B(t)) = \exp\!\bigl(\int_0^t \beta(s)\,d\mathcal{L}^1(s)\bigr)$, so \begin{align*} \phi(t) &\le \alpha\,\exp\!\Bigl(\int_0^t \beta(s)\,d\mathcal{L}^1(s)\Bigr) \quad \text{for every } t \in [0,T], \end{align*} which is the desired conclusion. [/guided] [/step]