of L1,loc for all p ≥ 1 . Every function belonging to , , where is an open subset of , is locally integrable. Proof. The case is trivial, therefore in the sequel of the proof it is assumed that . Consider the characteristic function of a compact subset of : then, for , where is a positive number such that = for a given is the Lebesgue measure of the compact set Then for any belonging to , by Hölder's inequality, the product is integrable i.e. belongs to and therefore Note that since the following inequality is true the theorem is true also for functions belonging only to the space of locally -integrable functions, therefore the theorem implies also the following result. . Every function in , , is locally integrable, i. e. belongs to . Note: If is an open subset of that is also bounded, then one has the standard inclusion which makes sense given the above inclusion . But the first of these statements is not true if is not bounded; then it is still true that for any , but not that . To see this, one typically considers the function , which is in but not in for any finite . L1,loc is the space of densities of absolutely continuous measures . A function is the density of an absolutely continuous measure if and only if . The proof of this result is sketched by . Rephrasing its statement, this theorem asserts that every locally integrable function defines an absolutely continuous measure and conversely that every absolutely continuous measures defines a locally integrable function: this is also, in the abstract measure theory framework, the form of the important Radon–Nikodym theorem given by Stanisław Saks in his treatise. Examples The constant function defined on the real line is locally integrable but not globally integrable since the real line has infinite measure. More generally, constants, continuous functions and integrable functions are locally integrable. The function for x ∈ (0, 1) is locally but not globally integrable on (0, 1). It is locally integrable since any compact set K ⊆ (0, 1) has positive distance from 0 and f is hence bounded on K. This example underpins the initial claim that locally integrable functions do not require the satisfaction of growth conditions near the boundary in bounded domains. The function is not locally integrable in : it