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Take a bounded set $S\subseteq \mathbb R^n$ with non-zero measure, and $M_S$ the set of measurable complex functions over $S$.

We know that the convergence in measure is metrizable and complete. Moreover the resulting distance is transational invariant, but it is not induced by a norm, since it doesn't respect the linearity.

As the title said, is this space a Frechet space? Or equivalently, is it locally convex?

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  • $\begingroup$ See en.wikipedia.org/wiki/… ... where your space is called $L^0$. It is an F-space (it is not locally convex, but it is a complete metric TVS). $\endgroup$ – GEdgar Mar 15 '18 at 13:04
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It is not locally convex.

Example. Let's take $L^0[0,1]$, equivalence classes mod null sets of measurable functions, with base for the neighborhoods of zero given by $$ V_\varepsilon = \big\{f \in L^0 : \lambda\{\,|f(x)| > \varepsilon\} < \varepsilon\big\} $$ for $\epsilon > 0$. I wrote $\lambda$ for Lebesgue measure. LINK

We claim that the constant $1$ belongs to the convex hull of every $V_\epsilon$. Fix $\varepsilon>0$. Choose $n \in \mathbb N$ with $\varepsilon > 1/n$. Then the following functions $f_{n,1},f_{n,2},\dots, f_{n,n}$ belong to $V_\epsilon$. Function $f_{n,j}$ is $n$ on the interval $$ \left[\frac{j-1}{n},\frac{j}{n}\right] $$ and $0$ elsewhere. The convex combination $$ \frac{1}{n}\sum_{j=1}^n f_{n,j} = 1 $$ is the constant $1$ a.e.

Any convex neighborhood of zero contains one of the sets $V_\varepsilon$, and therefore the constant $1$ belongs to it. The space is not locally convex.

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  • $\begingroup$ This is essentially the same as the proof that $L^p$, $0<p<1$, is not locally convex. $\endgroup$ – GEdgar Mar 15 '18 at 13:32

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