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Let $X$ be a complete separable metric space and $Y$ be a separable Banach space. We provide a proof of equivalence by linking explicitly the following statements:\\
\noindent \textbf{\textit{Lebesgue's Theorem.}} For every $\epsilon>0$ there exists a countable collection of closed sets $\left\lbrace C_n\right\rbrace $ of $X$ such that $$X=\bigcup_{n=1}^{\infty}C_n\;\;\text{and}\;\; \omega_f\left( C_n\right)<\epsilon\;\; \text{for each} \;\; n.$$
\textbf{\textit{Baire Characterization Theorem.}} For every nonempty perfect set $K\subset X$, the function $f|_K$ has at least one point of continuity in $K$. In fact, $C(f|_K)$ is dense in $K$.\\
\indent Moreover, replacing ``closed'' by ``open'' in the Lebesgue's Theorem, we obtain a characterization of continuous functions on space $X$.


Baire class one Lebesgue's theorem Baire Characterization Theorem

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How to Cite
Fenecios, J., & Racca, A. (2022). Equivalence of Lebesgue’s Theorem and Baire Characterization Theorem. Journal of the Indonesian Mathematical Society, 28(2), 158–163.


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