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I've found this problem while I was reading a paragraph about Riemann integration on some notes a mate gave me a long time ago.

Let $f \colon [a,b] \to \mathbb R$ be a bounded function. Suppose there exists a sequence of step functions $f_k$ s.t. $$ f(x)=\sup_{k \in \mathbb N} f_k(x) \quad \forall x \in [a,b] $$ Show that $f$ is lower semicontinuous in $[a,b]\setminus C$ where $C$ is a at most countable subset.

The problem seems quite easy: indeed, the pointwise sup of a family of lower semicontinuous functions is still semicontinuous (it's just a consequence of topology axioms, intersection of closed sets is closed). In other words, if $x$ is a point in which $f_k$ is l.s.c. for every $k \in \mathbb N$ so is $f$.

But what about the set $C$? Initially, I thought that the points of $C$ were the points of "jump" for $f_k$ (for some $k \in \mathbb N$). But is it true that a step function is not lower semicontinuous in a "jump" point? Is the problem clear? Hope so.

Thanks in advance.

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1 Answer 1

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The confusion might lie in what is meant by "step function." It is possible to have step functions that are lower semicontinuous, upper semicontinuous, or neither. For instance, if $f_k\colon[0,1]\to \mathbb{R}$ is the function $$f_k(x) = \begin{cases} 1-\frac{1}{k} & x\in[0,1/2].\\ 0 & x\in (1/2,1].\end{cases}$$ for each $k$, then the $f_k$ are upper semicontinuous, and their supremum is the function $f$ which is $1$ on $[0,1/2]$ and $0$ on $(1/2,1]$. This $f$ is upper semicontinuous but not lower semicontinuous. However, $f$ is lower semicontinuous on $[0,1]\smallsetminus \{1/2\}$.

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