not needed for the integral form. For a version of Grönwall's inequality which doesn't need continuity of and , see the version in the next section. Proof (a) Define Using the product rule, the chain rule, the derivative of the exponential function and the fundamental theorem of calculus, we obtain for the derivative where we used the assumed integral inequality for the upper estimate. Since and the exponential are non-negative, this gives an upper estimate for the derivative of . Since , integration of this inequality from to gives Using the definition of for the first step, and then this inequality and the functional equation of the exponential function, we obtain Substituting this result into the assumed integral inequality gives Grönwall's inequality. (b) If the function is non-decreasing, then part (a), the fact , and the fundamental theorem of calculus imply that Integral form with locally finite measures Let denote an interval of the real line of the form or or with . Let and be measurable functions defined on and let be a continuous non-negative measure on the Borel σ-algebra of satisfying for all (this is certainly satisfied when is a locally finite measure). Assume that is integrable with respect to in the sense that and that satisfies the integral inequality If, in addition, the function is non-negative or the function is continuous for and the function is integrable with respect to in the sense that then satisfies Grönwall's inequality for all , where denotes to open interval . Remarks There are no continuity assumptions on the functions and . The integral in Grönwall's inequality is allowed to give the value infinity. If is the zero function and is non-negative, then Grönwall's inequality implies that is the zero function. The integrability of with respect to is essential for the result. For a counterexample, let denote Lebesgue measure on the unit interval , define and for , and let be the zero function. The version given in the textbook by S. Ethier and T. Kurtz. makes the stronger assumptions that is a non-negative constant and is bounded on bounded intervals, but doesn't assume that the measure is locally finite. Compared to the one given below, their proof does not discuss the behaviour of the remainder . Special cases If the measure has a density with respect to Lebesgue measure, then Grönwall's inequality can be rewritten as If the function is