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.

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.