First version:

This is from an old script of my professor:

Let $f_n$ be a sequence of integrable functions. Let $f$ be a measurable function such that $$\lim_{n\to\infty} f_n(\omega) = f(\omega)$$ $P$-almost everywhere. Let $g\geq 0$ be a positive function with the property $$\int g\,dP<\infty$$ such that $$|f_n(\omega)|\leq g(\omega)$$ $P$-almost everywhere. Then $f$ is integrable and it is true that $$\lim_{n\to\infty}\int_\Omega f_n\,dP=\int_\Omega f\,dP$$

Second version:

I read this one on Wikipedia:

Let $f_n$ denote a sequence of real-valued measurable functions on a measure space $(\Omega ,\mathcal{A},P)$. Assume that the sequence converges pointwise to a function $f$ and is dominated by some integrable function $g$ in the sense that $|f_n(x)| \leq g(x)$ for all $x\in \Omega$. Then the limiting function $f$ is integrable and $$\lim_{n\to\infty}\int_\Omega f_n\,dP=\int_\Omega f\,dP$$

The difference here is that $f_n$ are real-valued measurable functions (not integrable as in the version above). Are these versions still equivalent?

Thanks for anyone who enlightens me.


Since each $f_n$ satisfies $|f_n(\omega)| \leq g(\omega)$ almost everywhere and $g$ is integrable, it follows from the monotonicity of the integral that each $f_n$ is integrable, so the second version follows from the first.

The first version follows from the second version by throwing away a null-set $N$ containing all points where $f_n$ doesn't converge to $g$.


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