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In Folland's book Real Analysis, he proves the following theorem (Theorem 2.27) concerning differentiation under integral sign.

Suppose $f:X \times [a,b] \to C$ is a complex valued function such that $f(\cdot,t)$ is integrable for each $t \in [a,b]$. Let $F(t)=\int_X f(x,t) d\mu(x)$. Then, if $\partial t f$ is dominated by an integrable function, we can differentiate $F(t)$ under the integral sign.

He then goes on to saying that if the parameter $t$ varies over a non-compact interval, we can still apply this theorem to any compact subinterval. But, my question is why is the theorem stated for compact interval $[a,b]$ in the first place? I do not see where/how the compactness of the interval is used in the proof

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Since differentiability is only a local property, you can replace $[a,b]$ by any open set, or $\mathbb{R}$. However, the key-point is that $\partial_t f$ must be dominated by an integrable function independent of $t$. In this formulation the interval $[a,b]$ indicates that this property only need to be satisfied locally.

If you replace $[a,b]$ by $\mathbb{R}$, this restriction is too strong: It is enough that for any open neighborhoud of any point $t$ there exists a uniformly integrable majorant.

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