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Let $V$ be the vector space of all real-valued Borel measurable functions on $[0,1]$. Show that convergence in measure (with respect to Lebesgue measure) is not given by a seminorm. That is, show that there no seminorm $\|\cdot\|$ on $V$ such that elements $f,f_1,f_2,\dots$ of $V$ satisfy $\lim_n \|f_n-f\| = 0$ iff $(f_n)$ converges to $f$ in measure.

(Hint: show that if such a seminorm exists, then for each positive $\epsilon$ there are functions $g_1, \dots, g_n \in V$ such that $\|g_i\| \leq \epsilon$ for each $i$ and such that $1/n \sum _{i= 1} ^n g_i $ is equal to the constant function 1.)

I have problem both showing the existence and then the contradiction. I'm grateful for hints or solutions.

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    $\begingroup$ The whole point is that existence is impossible; what you want to do is assume that such a seminorm exists and derive a contradiction, showing your assumption is impossible. I think there's a typo: "show that there seminorm" should be "show that there is no seminorm". $\endgroup$ Aug 5 '13 at 7:58
  • $\begingroup$ Yes there was a typo. Thanks. Can you help me further? $\endgroup$
    – Johan
    Aug 5 '13 at 10:47
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Only hints to give you, can provide details as you get further.

So first question, why is such a construction going to give you a contradiction? "Small pieces" whose average is the constant function $1$. What is a bound on the seminorm of this average function? What does it mean as you take $\epsilon$ towards $0$? You might consider explicitly choosing $\epsilon$ to be $1/k$ and looking at the resulting sequence $f_k$ (of average functions corresponding to $1/k$).

Second question towards construction: For simplicity, set $f=0$, we're converging in measure to $0$. Can you think of a canonical example of something that converges in measure to zero while only using indicators of intervals? Does this example change if you scale the sequence? (remember, you need to produce an average of things that give you the constant $1$).

(I should have added that the example I have in mind is $\chi_{[0,1]},\chi_{[0,1/2]},\chi_{[1/2,1]},\ldots$, so $\chi_{[0,1/n]}$ and $n$ of its shifts, increasing $n$. This way you can automatically get the seminorm of $\chi_{[0,1/m]}$ and $m$ of its shifts to all be less than a given $\epsilon$. This won't quite do, so you modify the example to scale by $n$, so you use $n \chi_{[0,1/n]}$ and $n$ of its shifts.)

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The sequence $a_n\chi_{(0,n^{-1})}$ converges in measure to $0$ for each $a_n\uparrow\infty$. This provides $n_0$ such that $\lVert \chi_{[0,n_0^{-1}]}\rVert=0$. Indeed, otherwise, take $a_n:=n\lVert \chi_{(0,n^{-1})}\rVert^{—1}$; then $\lVert a_n\chi_{(0,n^{-1})}\rVert=n$.

Hence there is function which is not $0$ almost everywhere, but the semi-norm which will do the job of this function is $0$. Since the condition on the semi-norm implies that each $f$ for which $\lVert f\rVert=0$ is $0$ almost everywhere, we get a contradiction.

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