# How to prove that a certain subset of $L_p$ space is closed?

For a probability measure space $(\mathbb{R}, \mathscr{B}, \mu)$, the spaces $L_p^{+}(\mu)$, $1 \leq p < \infty$, are defined to be the collection of all $\mu$-equivalence classes of $\mu$-measurable functions $f: \mathbb{R}_+ \rightarrow \mathbb{R}_+$ for which $\int_{\mathbb{R}_+} f^{p} \mathrm{d} \mu < \infty.$

If we define two subsets $E_c:= \{ f \in L^+_p(\mu) : \text{$f$is$\mu$-equivalent to a real-valued non-negative function on$\mathbb{R}_+$that is concave} \}$, $E_m:= \{ f \in L^+_p(\mu) : \text{$f$is$\mu$-equivalent to a real-valued non-negative function on$\mathbb{R}_+$that is monotone increasing} \}$,

are these two subsets $E_c$ and $E_m$ closed in $L^+_p$ space?

I thought that this subset $E$ of the $L_p$ space should be closed.

I'm curious to know how to show it rigorously. Could anyone help to show it please? Thanks in advance:)

## 1 Answer

Since these are not functions but equivalence classes of functions, your definition doesn't quite make sense. Perhaps you mean

$f$ is $\mu$-equivalent to a function on $\mathbb R$ that is concave and non-decreasing

But this doesn't quite work either. Let $\mu$ be Lebesgue measure on $[0,1]$ and consider the functions $f_n(x) = \sqrt{x+1/n}$ on $[0,1]$. These are in your set, and converge in $L_p$ to $f(x) = \sqrt{x}$. But that is not in your set, because it is not $\mu$-equivalent to a concave function on $\mathbb R$. It is concave on $[0,1]$, but can't be extended to negative reals and remain concave.

You might want to restrict the concavity requirement to the support of $\mu$, but that is not necessarily a convex set, and I don't know how you want to define concavity for functions whose domain is not convex.

• Thank you for your answer. I'll think about it more carefully:) – Paradiesvogel Mar 17 '17 at 5:53
• When we say a concave function, its domain should be a convex set. Otherwise, it doesn't make sense:) – Paradiesvogel Mar 28 '17 at 23:09