Show that the following f is integrable on every Jordan bounded set? (proof verifying) Q: let $f:\mathbb{R}^n\to \mathbb{R}$ satisfy that for each open ball $U$ then $f$ is integrable (such that $\int|f|<\infty$)) over $U$ and $\int_{U}f=v(U)$ where v is the volume of $U$. Show that for every Jordan bounded set $E$ it is that $f$ is integrable over $E$ and $\int_{E}f=v(E)$

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*Jordan bounded sets are sets with negligible boundary.

Hey, I would really like to know if the following proof holds:
Let $E$ be a bounded Jordan set. for each $x\in E$ denote $B_x=B(x,r_x)$ where $r_x$ will be stated later.
then $\bigcup_{x\in E}B_x\supseteq E$.
Now, I claim that $ \{x\in E: \forall r. B(x,r)\cap E^\circ \neq \emptyset \land B(x,r)\cap E^{exterior}\neq \emptyset\}$ in negligible because each x that satisfies that belongs to $\partial E$ and hence this group is subset of negligible group and hence negligible itself.
Hence, for almost every x (except negligible set) we can choose $r_x$ such that $B_x\subseteq E$ and hence $\bigcup_{x\in E\\x\notin \partial E}B_x\subseteq E$ and therfore $$\int_{\bigcup_{x\in E\\x\notin \partial E}B_x}f = \int_Ef =(*)$$ and because $\int|f|<\infty$ over every open ball, then we got integrablity of $f$ over $E$
Second part: I claim that for each open set $U$ then $\int_Uf=v(U)$ because we can take an open ball $B$ such that $U\subseteq B$ and define $\tilde{f} = \cases{f\;\;x\in U\\0\;\;x \in B/U}$ and hence it is that $\int_B\tilde{f}=\int_Uf=v(U)$
Now we say that $(*)=v(\bigcup_{x\in E\\x\notin \partial E}B_x) = v(E)$ because the difference between the sets is negiligible.
Is my proof right? I am especially wondering if the claim in the second part is actually true and well explained?
Appreciate any help!
 A: Formal solution of what I have pointed in the comments:
Proof of the integrability of $f$:
claim: $\mathcal{B}_f$ (denotes the set of points where $f$ is not continous) is negligible.
proof: Indeed, $\mathcal{B}_f = \bigcup_{i=1}^{\infty}\mathcal{B}_{f|_{B(0,i)}}$ is a negligible set as $f$ is integrable over each open ball (corollary from Lebesgue sentence).

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*remark: The above proof does not assume that countable union of negligible sets is negligible (it isn't necessarily), but that for each $i$ the set $\mathcal{B}_{f|_{B(0,i)}}$ is negligible, and each Jordan set $E$ is contained in such ball

corollary: Again, from Lebesgue $f$ is integrable over every bounded Jordan set.
Proof of the property of $f$:
claim: $f|_{\mathbb{R}^n\setminus\mathcal{B}_f}=1$
proof: Otherwise, there is continuity point ,$x$ of $f$ which satisfies (WLOG) that $f(x)>1$. Hence, there is an open neighborhood, $U_x$ of $x$ where $f(U_x)>1$ and as there can be found a ball, $B(x,\delta_{U_x})$ contained in $U_x$ by the linearity of Integrals: $\int_{B(x,\delta_{U_x})}f>\int_{B(x,\delta_{U_x})}1=vol(B(x,\delta_{U_x}))$ which contradicts the given.
corollary: $\int_E{f} = \int_{E\setminus\mathcal{B}f|_E}1+\int_{\mathcal{B}f|_E}f = vol(E)$
