The strategy is to multiply the ODE by a function $\mu(t)$ chosen so that the left-hand side becomes the derivative of a product, then integrate directly. The proof has three parts: construction of the integrating factor, derivation of the solution formula, and uniqueness.

**Step 1: Construction of the integrating factor.** Define $\mu: I \to \mathbb{R}$ by \begin{align*} \mu(t) &:= \exp\!\Bigl(\int_{t_0}^t a(s)\,d\mathcal{L}^1(s)\Bigr). \end{align*} Since $a$ is continuous, the integral $\int_{t_0}^t a(s)\,d\mathcal{L}^1(s)$ is $C^1$ in $t$ by the fundamental theorem of calculus, so $\mu \in C^1(I)$. Moreover $\mu(t) > 0$ for all $t \in I$ (as the exponential of a real number), and $\mu(t_0) = e^0 = 1$. Differentiating: \begin{align*} \mu'(t) &= a(t)\,\exp\!\Bigl(\int_{t_0}^t a(s)\,d\mathcal{L}^1(s)\Bigr) = a(t)\,\mu(t). \end{align*}

**Step 2: Derivation of the solution.** Suppose $y: I \to \mathbb{R}$ is a $C^1$ solution of $y' + a\,y = b$. Multiply both sides by $\mu(t)$: \begin{align*} \mu(t)\,y'(t) + \mu(t)\,a(t)\,y(t) &= \mu(t)\,b(t). \end{align*} By Step 1, $\mu(t)\,a(t) = \mu'(t)$, so the left-hand side is $\mu(t)\,y'(t) + \mu'(t)\,y(t) = \frac{d}{dt}\bigl[\mu(t)\,y(t)\bigr]$ by the product rule. Hence \begin{align*} \frac{d}{dt}\bigl[\mu(t)\,y(t)\bigr] &= \mu(t)\,b(t). \end{align*} Integrating both sides from $t_0$ to $t$ and using $\mu(t_0)\,y(t_0) = 1 \cdot y_0 = y_0$: \begin{align*} \mu(t)\,y(t) - y_0 &= \int_{t_0}^t \mu(s)\,b(s)\,d\mathcal{L}^1(s). \end{align*} Dividing by $\mu(t) > 0$ gives the claimed formula: \begin{align*} y(t) &= \mu(t)^{-1}\Bigl(y_0 + \int_{t_0}^t \mu(s)\,b(s)\,d\mathcal{L}^1(s)\Bigr). \end{align*} This shows that every solution must equal the right-hand side, establishing uniqueness. It remains to verify existence: define $y$ by this formula. Then $y \in C^1(I)$ (since $\mu$ and $b$ are continuous), $y(t_0) = \mu(t_0)^{-1}(y_0 + 0) = y_0$, and reversing the computation above confirms that $y' + a\,y = b$.