Theorem 3

For a vector function $f$ defined on an interval $I$, with values in a complete normed space $E$ over $\mathbf{R}$, to be regulated, it is necessary and sufficient that it have a right limit and a left limit at every interior point of $I$, and a right limit at the left-hand endpoint of $I$ and a left limit at the right-hand endpoint of $I$, when these points belong to $I$. The set of points of discontinuity of $f$ in $I$ is thus countable.

Since every interval is a countable union of compact intervals, one can restrict oneself to proving th. 3 when I is compact, say I = [a, b].

1 The condition is necessary. Suppose that f is regulated and let x be a point of I different from b. By hypothesis, for every ε > 0 there is a step function g such that \|f(z) - g(z)\| \leq \varepsilon for every z \in I; since g has a right limit at the point x there exists a y such that x < y \leq b and such that, for every pair of points z, z' in the interval ]x, y] one has \|g(z) - g(z')\| \leq \varepsilon, and in consequence \|f(z) - f(z')\| \leq 3\varepsilon; this proves (by Cauchy’s criterion) that f has a right limit at the point x. In the same way one proves that f has a left limit at every point of I different from a.

2° The condition is sufficient. Suppose it is satisfied; for each x \in I there is an open interval V_x = ]c_x, d_x[ containing x and such that on the intersection of I with each of the open intervals ]c_x, x[, ]x, d_x[ (when the intersection is not empty) the oscillation of f is \leq \varepsilon. Since I is compact there is a finite number of points x_i in I such that the V_{x_i} form a cover of I; let (a_k)_{0 \leq k \leq n} be the sequence obtained by arranging in increasing order the points of the finite set formed by the points a, b and those points x_i, c_{x_i} and d_{x_i} which belong to I; each of the intervals ]a_k, a_{k+1}[ (0 \leq k \leq n - 1) being contained in an interval ]c_{x_i}, x_i[ or ]x_i, d_{x_i}[, the oscillation of f there is \leq \varepsilon; let c_k be one of the values of f on ]a_k, a_{k+1}[; on putting g(a_k) = f(a_k) for 0 \leq k \leq n, and g(x) = c_k for all x \in ]a_k, a_{k+1}[ (0 \leq k \leq n - 1), one defines a step function g such that \|f(z) - g(z)\| \leq \varepsilon on I; so f is regulated on I.

Finally, let us show that if f is regulated on I then its set of points of discontinuity is countable. For every n > 0 there exists a step function g_n such that \|f(x) - g_n(x)\| \leq 1/n on I; since the sequence (g_n) converges uniformly to f on I, we see that f is continuous at every point where the g_n are all continuous (Gen. Top., X, p. 281, cor. 1); but since g_n is continuous except at the points of a finite set H_n it follows that f is continuous at the points of the complement of the set H = \bigcup_{n} H_n, which is countable.