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

In my lecture notes there is the following exercise:

"Characterize those measure spaces $(X, B, \mu)$ on which the semi-norm $\|f\| = \int_X |f| d \mu$ defined on $L^1(X) = \{ f \mid f \text{ measurable and} \int_X |f| d \mu < \infty \}$ of Lebesgue-integrable functions is a norm."

I thought that if I take $X$ to be finite and $\mu$ to be the counting measure then $\|\cdot\|_1$ is a norm. But I think the exercise asks me to use the Lebesgue measure so $\mu(X) = 0$ if $X$ is finite and my example breaks.

What's the correct answer?

share|cite|improve this question
I don't think Lebesgue-integrable means anything else than $\mu$-integrable here. This is sometimes called "integrable in the sense of Lebesgue". Also: you don't characterize anything, you only give an example. A hint: what happens if there's a nonvoid $\mu$-null set? – t.b. Jul 8 '12 at 20:03
@DavideGiraudo The integral over the whole space, that was a typo. – Rudy the Reindeer Jul 8 '12 at 20:25

Assume that $\lVert \cdot \rVert$ is a norm. If $N$ is a measurable set of measure $0$, then $\int_X\chi_Nd\mu=0$ hence $\chi_N$ is identically $0$ and $N$ is the emptyset. So each non-empty measurable set set has a positive measure.

Conversely, if each non-empty measurable set has a positive measure, let $f$ a function such that $\lVert f\rVert=0$. Then $$0=\lVert f\rVert\geq\int_{\{f\geq 2^{-n}\}}|f|d\mu\geq 2^{—n}\mu(\{|f|\geq 2^{-n}\})$$ hence $\mu(\{|f|\geq 2^{-n}\})=0$ and $\mu(\{|f|>0\})=0$. By hypothesis, it implies that $\{|f|>0\}$ is empty, hence $f=0$.

In general, to define a norm by this way, we rather consider the equivalence classes of functions. Otherwise, we may have some problems. For example, if $\mathcal B$ is the smallest $\sigma$-algebra containing all the open sets of a Hausdorff topology, each singleton is measurable and should have a positive measure. So $X$ can only be at most countable.

share|cite|improve this answer

Here is a large source of them. Suppose $\Omega$ is a set that is partitioned by subsets $$\{E_n\}_{n=1}^\infty$$
Suppose that $\mu$ is a measure on the $\sigma$ algebra $\mathcal{S}$ generated by the $E_k$ and that $\infty > \mu(E_k) > 0$, $1\le k \le n$. Then $f\mapsto \int_{\Omega} |f|\,d\mu$ is a norm on $L^1(\Omega, \mathcal{S}, \mu)$.

Some variation on this is probably your complete answer. Note that any $\mu$-null set that is nonvoid does you in on the spot.

share|cite|improve this answer

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.