The indefinite integrals are the continuous functions. If f(x) is a bounded function, this is obvious by virtue of theorem of finite increment. Let us assume, then, f(x) is summable but unbounded; then, we can find N sufficiently large such that the integrals of f(x) over both sets \(E(f > N)\) and \(E(f < -N)\) are both less than \(\varepsilon \) in modulus. Let us set \(f = f_1 + f_2,\) where \( f_1\) is zero on both sets \(E(f > N)\) and \(E(f < -N)\) and \(f_2\) is zero on the set \(E(-N \le f \le N)\) . Then the indefinite integral of \(f_1\) is a continuous function; and the integral of \(f_2\) in any interval being \(2\varepsilon \) at most, around any point \(x_0\) , therefore, we can find an interval in which the increment of f(x) is \(3\varepsilon \) at most. This proves that f(x) is continuous.