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I am reading something on BV functions and I am totally stucked with some assertion made during the development of one proof. How can I prove the following?

Let $\varOmega \subset \mathbb{R}^n$ be an open set and $\{E_j\}_{j \in \mathbb{N}} \subset \mathbb{R}^n$ be a sequence of measurable sets such that $\chi _{E_j} \rightarrow f \in BV(\varOmega)$. Then $f$ is the characteristic function of some set $E$ in almost every point. (where $\chi_{A}$ denotes the characteristic function of $A$)

J. Salieri
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1 Answers1

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If $\chi_{E_j}$ converges to $f$ in $L^1(\Omega)$ then you can extract a subsequence $\chi_{E_{j_k}}$ which converges to $f$ pointwise a.e. But then if $\chi_{E_{j_k}}(x)\to f(x)$, since $ \chi_{E_{j_k}}(x)$ takes only values $0$ and $1$, for all $k$ large you have that either $\chi_{E_{j_k}}(x)=0$,in which case $f(x)=0$, or $\chi_{E_{j_k}}(x)=1$ for all $k$ large, in which case $f(x)=1$. Thus, at every point where you have pointwise convergence $f$ is either zero or 1

Gio67
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