Example of a measure which is not an $L^1$ function.

Question

If $f\in L^1(\Omega)$, then $f$ defines a measure on $\Omega$. I wonder, can a function $f\not\in L^1(\Omega)$ define a finite measure on $\Omega$?

I know that Dirac delta defines a measure which is not in $L^1$, but I am bit confused whether it is a function, because it is not defined on $\Omega$ but on $\mathcal{C}_c^\infty(\Omega)$ (the space of compactly supported smooth functions) . Thanks!!

@Martín-BlasPérezPinilla $\Omega \subset \mathbb{R}$ , with Lebesgue measure. — Arun, May 08 '18 at 10:08
@tomasz If $f\in L^1$, then $v_f (A)$ as in Motyla's answer defines a measure on $\Omega$. — Arun, May 08 '18 at 10:16
You want a measure that is not "absolutely continuous". Probabably any measure theory text has discussion of this. — GEdgar, May 08 '18 at 10:40
@Arun: I know that, I'm asking what you mean by a function which is a measure, but is not in $L^1$. In what sense would the function be a measure? — tomasz, May 08 '18 at 14:27
If the function is non-negative, then all you really need is for it to be measurable, it's just that if it's not in $L^1$, the measure will be infinite. — tomasz, May 08 '18 at 14:29

score 2 · Answer 1 · answered May 08 '18 at 09:55

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To every measurable function $f:\Omega \to \mathbb{R}$ we can associate an expression $$v_f (A) =\int_A f(t) l_n (dt ) $$ where $l_n $ is $n$-dimensional lebesgue measure. If $v_f (A) $ is finite for every lebesgue measurable subset of $\Omega$ then it is easy to show that $f\in L_1 (\Omega ).$

answered May 08 '18 at 09:55

Thanks. So you mean every measure which are generated by a function is in $L^1$? – Arun May 08 '18 at 10:02
@Arun: The right way of saying that is "every measure that is absolutely continuous with respect to the Lebesgue measure is in $L^1$". You don't say that a measure is "generated by a function", even if it is clear what you mean. – Giuseppe Negro May 08 '18 at 10:24
@GiuseppeNegro Thanks for correcting me. Can you give me some reference for the result you stated. It is too much of asking, but I just wonder is there any function space bigger than $L^1$? – Arun May 08 '18 at 10:29

score 1 · Accepted Answer · answered May 08 '18 at 10:49

The problem is that you are unsure on the identification between (some) functions and measures. Given a function $f\in L^1(\mathbb R^n)$, you can define a measure $\mu_f$ by $$ \mu_f(A):=\int_{\mathbb R^n} f(x)\, dx, \quad \forall A\ \text{Lebesgue measurable},$$ or, more shortly, $d\mu_f := f\, dx$. The general theorem that governs this kind of constructions is the Radon-Nikodym theorem, but you do not need its full force, only the statement. If a measure $\mu_f$ is constructed like that, you say that it is absolutely continuous with respect to the Lebesgue measure $dx$, and you say that $f$ is the Radon-Nikodym derivative of $\mu_f$ with respect to $dx$, which alludes to the formal manipulation $$ d\mu_f = f\, dx \quad \overset{\text{formal}}{\iff}\quad f=\frac{d\mu_f}{dx}.$$

Your question is now: can we do this construction with a function $f$ that does not belong to $L^1(\mathbb R^n)$? The answer is no: the writing $$\tag{!!} \int_A f\, dx $$ does not make sense for all measurable $A\subset \mathbb R^n$ unless $f\in L^1(\mathbb R^n)$.

score 0 · Answer 3 · answered May 08 '18 at 10:10

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In the whole space, $f = 1\not\in L^1$ as measure is the Lebesgue measure.

answered May 08 '18 at 10:10

Martín-Blas Pérez Pinilla

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But $f$ does not define a finite measure. Could we expect a finite measure not in $L^1$. – Arun May 08 '18 at 10:24
@Arun, then edit your question an add the condition. – Martín-Blas Pérez Pinilla May 08 '18 at 10:29
@Arun, what means that a measure is in $L^1$? – Martín-Blas Pérez Pinilla May 08 '18 at 10:30
I have edited the question. I mixed up everything. Sorry for that. – Arun May 08 '18 at 10:40

Example of a measure which is not an $L^1$ function.

3 Answers3