Prove that: Suppose $f \colon \Bbb R \rightarrow \Bbb R $, where $f$ is continuous on $(-\infty, 0)$ and on $(0,\infty)$. Show that $f$ is measurable.

Can someone please help me with this proof? My TA told me not to look at it in my self study class because we are not going to cover this. I am trying to understand this although my TA told me not to.

Can someone please give a proof for this. This seems like a good proof to know about continuous and measurbale functions which I might learn later on in another class.

  • $\begingroup$ Well, group theory seems irrelevant to the topic? $\endgroup$ – awllower Apr 15 '13 at 12:54

continuous a.e. functions are Lesbesgue-measurable. More generally, if $f = g$ $\mu$-a.e. then $g$ is measurable, given that $\mu$ is complete.

Since $f − g = 0$, μ-a.e., $(f − g)^{−1}(0) = X −N$, where $N = f^{−1}(R−0)$ is some null set of $M$ (since $0$ is Borel), giving us that $(f − g)^{−1}(0)$ is measurable. Take any Borel subset $B$ of $\mathbb{R}$ now. Then $(f − g)^{−1}(B) = (X − N) \cup F_1$ or $ = F_2$, depending on whether $B$ contains zero, where $F_1, F_2$ are null sets contained in $N$. Again by the completeness of $μ$ this set is measurable. So $f − g$ is measurable. Now $g = f −(f −g)$, thus is measurable.


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