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Let $f$ a real function Lebesgue measurable over $\mathbb R^{k}$. Show that exists borelians functions $g$ and $h$ such that $g(x) = h(x)$ for $m$-almost everywhere point $x$ and $g(x)\leq f(x)\leq h(x)$ for all $x\in \mathbb R^{k}$.

My answer: I use the Lusin's Theorem, I supusse that exists $A \subset X$, measurable set, that $\mu(A)< +\infty$ and get a compact set $K$ such that

$$ f|_K $$ is continuous

Next $f$ is borelian, then we can extend this function to all $\mathbb R^{k}$ with partition unit, remains to get $g$ and $h$.

I think that \begin{eqnarray} g(x) &=& f(x)+1 \;\;\;\; x\in A-K\\ & & f(x) \;\;\;\;\;\;\;\;\;\; x \in K \end{eqnarray}


\begin{eqnarray} h(x) &=& f(x)-1 \;\;\;\; x\in A-K\\ & & f(x) \;\;\;\;\;\;\;\;\;\; x \in K \end{eqnarray}

Because $g$ and $h$ are different in a set of measure $0$ because the Lusin's theorem, and then $g=h$ almost everywhere.

It is convincing?

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