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I have just started with Measure Theory and I have read several times that

Every locally integrable function defines a Radon measure.

I understand this statement in the sense that if we have $f\in L^1_{loc}(\mathbb{R}^N)$ (respecto to Lebesgue measure) , then $\mu(E)=\int_{E}f(x)dx$ (where the integral is respect to Lbesgue measure) defines a Radon measure. Is this right? How could I prove it? Is there any measure which is not of this form?

Maybe these questions are trivial, but I am a little lost. Thanks.

Edit: After see the answer of @JustDroppedIn I was wondering what happens if $f$ is not non-negative. I thought that in this case we would obtain a signed measure. However, if $f$ we descompose $f=f^+-f^-$, where $f^+=\frac{f+|f|}{2}\geq{0}$ and $f^-=\frac{f-|f|}{2}\geq{0}$, then $\mu(E)=\int_E f=\int_E f^+-\int_E f^- $ and, by the answer of @JustDroppedIn, $\mu$ is the difference of two positive Radon measures. The problem is that this difference doesn't have to bee a signed measure so I would like to know what extra hypothesis we neeed above $f$ to obtain a signed Radon measure.

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  • $\begingroup$ Have you tried checking the definition of a Radon measure? $\endgroup$ Commented Jun 27, 2020 at 10:32
  • $\begingroup$ Yes, I tried it. But I don't know how to start $\endgroup$ Commented Jun 27, 2020 at 10:37
  • $\begingroup$ What are the requirements for a measure to be Radon? $\endgroup$ Commented Jun 27, 2020 at 10:56
  • $\begingroup$ The delta measure at $0$ defined by $\delta (A)=1$ if $0 \in A$ and $0$ if $0 \notin A$ is a measure which is not of this form for any $f$. $\endgroup$ Commented Jun 27, 2020 at 12:00

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Let's recall: a Radon measure is a measure $\mu:\mathcal{B}(\mathbb{R}^N)\to[0,\infty]$ such that $\mu$ is regular and $\mu$ is locally finite. Suppose that $f\in L^1_{loc}(\mathbb{R}^N)$ and that $f\geq0$. Define $\mu(E)=\int_Ef$. This is a measure and the fact that $f\in L^1_{loc}$ shows immediately that $\mu$ is locally finite. Regularity of $\mu$ follows directly from the regularity of the Lebesgue measure and an application of the well-known convergence theorems.

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  • $\begingroup$ Can we extend this result to signed measures, that is $\mu:\mathcal{B}(\mathbb{R}^N)\to \mathbb{R}$? $\endgroup$ Commented Jun 27, 2020 at 11:08
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    $\begingroup$ @mathlife I believe yes. You just have to be careful that $\mu$ omits either $+\infty$ or $-\infty$. $\endgroup$ Commented Jun 27, 2020 at 11:20
  • $\begingroup$ .@JustDroppedIn Hello, I was reading your answer again and I was wondering why we must omit the case $\pm \infty$ when $\mu$ is signed but we don't have that omit the case $\infty$ when $\mu$ is a positive meausre $\endgroup$ Commented Jun 30, 2020 at 11:36
  • $\begingroup$ @mathlife It is by the definition of signed measures. A signed measure by definition must NOT attain both values $\pm\infty$. There is no such restriction on positive measures of course. If you want, we can discuss on why signed measures are defined in this way, but this is another topic. You can check Folland's text. It is the best thing when it comes to measure theory. $\endgroup$ Commented Jun 30, 2020 at 12:03
  • $\begingroup$ .@JustDroppedIn For example in Definition 4.1.1 of this text math.uwaterloo.ca/~beforres/PMath451/Course_Notes/Chapter4.pdf there is not such restriction about values $\pm \infty$. Essentially my new question can be reformulated as: What happens if $f$ is not a positive function? $\endgroup$ Commented Jun 30, 2020 at 14:25

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