On any bounded parameter interval $|t|\le T$, the derivative of the integrand has the integrable majorantThe dominated convergence theorem therefore justifies differentiation under the integral sign at every real $t$, givingUse integration by parts in this absolutely convergent integral, with $x{e}^{-x^2}dx=-\frac{1}{2}d(e^{-x^2})$. Both boundary terms vanish, since the sine is zero at $x=0$ and the Gaussian tends to zero at infinity. HenceIt follows that $f^{\prime }=-4t^{3}f$, or equivalently $(e^{t^4}f(t){)}^{\prime }=0$. This identity holds on all of $\mathbb{R}$, including zero. With the supplied Gaussian integral value,The fourth power comes from the printed phase $2x{t}^2$; the parameter has been differentiated with that phase throughout.

A sigma-algebra $\mathcal{F}$ on $\Omega$ contains $\Omega$, is closed under complements relative to $\Omega$, and is closed under countable unions. Consequently it contains the empty set and is closed under countable intersections. A measure space is a triple $(\Omega ,\mathcal{F},\mu )$ with a map $\mu :\mathcal{F}\to [0,\infty ]$ satisfying $\mu (\varnothing )=0$ and countable additivity on pairwise disjoint measurable sets.

For the increasing sequence, set $C_1=A_1$ and $C_n=A_n\setminus A_{n-1}$ for $n\ge 2$. These measurable sets are disjoint, their union is $A$, and $A_N={\bigcup }_{n=1}^N{C}_n$. Thus countable additivity and finite additivity giveThe identity permits the value infinity; it uses no subtraction of infinite measures. This is continuity from below of a measure.

For a decreasing sequence, the analogous assertion is not always true. On $\Omega =\mathbb{N}$ with the counting measure, take $B_n=\{n,n+1,\dots \}$. Then ${\bigcap }_n{B}_n=\varnothing$, so $\mu (B)=0$, whereas $\mu (B_n)=\infty$ for every $n$ and their limit is infinity.

A sufficient extra condition for continuity from above of a measure is $\mu (B_1)<\infty$. Indeed $B_1\setminus B_n$ increases to $B_1\setminus B$, and the increasing-sequence result givesMore generally it suffices that one of the $B_n$ has finite measure. The missing finiteness assumption is exactly what the counterexample violates.