Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Lemma: If $f$ is a nonnegative measurable function and $\int f<\infty$, then $\int f = \lim_{n\to \infty}\int_{[-n,n]} f$.

Proof: We know that $\int_{[-n,n]}f = \int f \cdot \chi_{[-n,n]}$. If $n < m$, then $f \cdot \chi_{[-n,n]} \leq f \cdot \chi_{[-m,m]} \leq f$. So the limit exists and $\lim_{N\to \infty} \int_{[-N,N]} f\leq \int f$.

Now we want to show the reverse. Suppose $\epsilon > 0$. There exists a simple integrable function $g$ such that $0 \leq g \leq f$, and $\int f \leq \int g + \epsilon$. Let $n$ be sufficiently large such that $g = 0$ outside of $[-n,n]$. Then, $g = g \cdot \chi_{[-n,n]}$, so $\int g = \int_{[-n,n]} g$. Now $g\leq f$ implies $g \cdot \chi_{[-n,n]} \leq f \cdot \chi_{[-n,n]}$. Therefore, $\int f \leq \int g + \epsilon = \int_{[-n,n]}g + \epsilon \leq \int_{[-n,n]}f + \epsilon$ for $n > N \in \mathbb{N}$. So $\int f \leq \lim_{n\to \infty} \int_{[-n,n]} f + \epsilon$. Now letting $\epsilon \to 0$, we obtain $\int f \leq \lim_{n\to \infty} \int_{[-n,n]} f $, and we are done.

There is a problem in this proof, but I can't seem to find/correct it. Can anyone help?

share|cite|improve this question
up vote 1 down vote accepted

A simple function is a linear combination $\sum\limits_{k=1}^na_k\mathbf 1_{A_k}$ of indicator functions $\mathbf 1_{A_k}$ of measurable sets $A_k$ but one does not usually asks that the sets $A_k$ are intervals nor even bounded, only measurable. Hence the step Let $N$ be sufficiently large that $g=0$ outside of $[-N,N]$ is not guaranteed.

To fix the solution proposed by the OP, choose $N$ such that the integral of $g−g\mathbf 1_{[−N,+N]}$ is at most $\epsilon$, and proceed.

share|cite|improve this answer
How would I fix the solution? – Cardflow Mar 17 '12 at 6:55
Choose $N$ such that the integral of $g-g\mathbf 1_{[-N,+N]}$ is at most $\epsilon$, and proceed. – Did Mar 17 '12 at 6:59

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.