两端对 $x$ 求导，解微分方程得 $f(t)=A(t)\exp\left(\int_{t}^{x} g(r) , \mathrm{d}r\right)$ 代入原不等式得证

核心构造函数 $v(t)=\exp\left(-\int_{a}^{t} g(r) , \mathrm{d}r\right)\int_{a}^{t} f(r)g(r) , \mathrm{d}r$

则 $$ \begin{align} v’(t) & =\left(f(t)-\int_{a}^{t} f(r)g(r) , \mathrm{d}r \right)g(t)\exp\left(-\int_{a}^{t} g(r) , \mathrm{d}r \right) \ & \le A(t)g(t)\exp\left(-\int_{a}^{t} g(r) , \mathrm{d}r \right) \end{align} $$ 两边对 $[a,x]$ 求定积分，根据 $v(a)=0$ 得 $$ \begin{align} v(x) & \le \int_{a}^{x} A(t)g(t)\exp\left(-\int_{a}^{t} g(r) , \mathrm{d}r \right) , \mathrm{d}t \ \int_{a}^{x} f(t)g(t) , \mathrm{d}t & \le \int_{a}^{x} A(t)g(t)\exp\left(\int_{a}^{x} g(r) , \mathrm{d}r - \int_{a}^{t} g(r) , \mathrm{d}r \right) , \mathrm{d}t \ \int_{a}^{x} f(t)g(t) , \mathrm{d}t & \le \int_{a}^{x} A(t)g(t) \exp\left(\int_{t}^{x} g(r) , \mathrm{d}r \right) , \mathrm{d}t
\end{align} $$ 从而轻易有 $$ f(x) \le A(x) + \int_{a}^{x} f(t)g(t) , \mathrm{d}x \le A(x) + \int_{a}^{x} A(t)g(t)\exp\left(\int_{t}^{x} g(r) , \mathrm{d}r \right) , \mathrm{d}t
$$