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!!

Answer 1

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 $$ d\mu_f = f\, dx \quad \overset{\text{formal}}{\iff}\quad f=\frac{d\mu_f}{dx}.$$

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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 $f\in L^1(\mathbb R^n)$.

Answer 2

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 $f\in L_1 (\Omega ).$

Answer 3

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