Proposition 8

If f is continuous on I × A one has

("formula for interchanging the order of integration").

We shall show that, for all $y \in A$, one has

Since the two sides of (16) are functions of $y$, and equal for $y = c$, it will suffice to prove that they are differentiable on $]c, d[$ and that their derivatives are equal at every point of this interval. If one puts $g(\alpha) = \int_a^b f(x, \alpha) dx$, and $h(x, y) = \int_c^y f(x, \alpha) dx$, the relation (16) can be written

Now, the derivative of the first term with respect to $y$ is $g(y)$, while that of the second is $\int_a^b h_y'(x, y) dx$, by II, p. 74, corollary, since $h_y'(x, y) = f(x, y)$ is continuous on $I \times A$; the two expressions thus obtained are identical.

Suppose now that $A = [c, d]$ is a compact interval in $\mathbf{R}$, and I an arbitrary interval in $\mathbf{R}$; let $f$ be a continuous function on $I \times A$, with values in E, such that the integral $g(\alpha) = \int_I f(t, \alpha) dt$ is convergent for all $\alpha \in A$; even if $g(\alpha)$ is continuous on $A$ one cannot always interchange the order of integration in the integral $\int_c^d d\alpha \int_I f(t, \alpha) dt$, for the integral $\int_I dt \int_c^d f(t, \alpha) d\alpha$ may not exist, or it may be different from the integral $\int_c^d d\alpha \int_I f(t, \alpha) dt$ (cf. II, p. 87, exerc. 7). One has, however, the following result: