# A question on measure theory

Hi: I need assistance with the following problem:

Let $(X,M,\mu)$ be a measure space. Let $X$ be the union of a countable ascending sequence of measurable sets $\{X_n\}$ and $f$ a nonnegative measurable function on $X$. Show that $f$ is integrable over $X$ if and only if there is an $N \geqslant 0$ for which $\int_{X_n} f~\text{d}\mu \leqslant N$ for all $n$.

I think I have to use these facts: Since $X_n\subset X_{n+1}$ and $X=\cup X_n$, $\mu(X) = \lim X_n$. For one direction, I know I have to show $\int_X fd\mu <\infty$.

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Try proving the contrapositive. It should pop right out. –  Brett Frankel Mar 8 '12 at 0:32
($\Rightarrow$) is obvious, just use $N = \int_X f\ d\mu$. ($\Leftarrow$) is less obvious. Set $f_n = f$ on $X_n$ and $0$ elsewhere. Because $X_n$ are measurable, $f_n$ is also measurable and nonnegative, and then you could just use Fatou's lemma. –  dtldarek Mar 8 '12 at 0:43
@dtldarek: how do I use the condition that $X=\cup X_n$? –  Jacob Mar 8 '12 at 1:28
@Jacob You need $\lim_n f_n = f$, and for that both $X_n \subset X_{n+1}$ and $X = \bigcup_n X_n$. –  dtldarek Mar 8 '12 at 8:15

If $f$ is integrable, then $$\int_{X_n}f\,d\mu\leq\int_{X}f\,d\mu$$ for all $n$, and we can take $N$ to be $\int_Xf\,d\mu$.
Conversely, if $\int_{X_n}f\,d\mu\leq N$ for all $n$, we consider the monotone sequence (this is where we use $X_n\subset X_{n+1}$) of functions $\{f\, 1_{X_n}\}_n$. This sequence converges pointwise to $f$ (this is where we use that $\cup X_n=X$), so by the Monotone Convergence Theorem, $$\int_Xf\,d\mu=\lim_{n\to\infty}\int_{X}f\,1_{X_n}\,d\mu=\lim_{n\to\infty}\int_{X_n}f\,d\mu\leq N$$