Defining a probability measure over a training sample? I am reading a paper which considers Lasso from the viewpoint of Robust Optimisation. In section 4, the authors begin to discuss the statistical consistency of Lasso. They introduce the a set $P_{n}$ in the following proposition

I am struggling to understand what is meant by this definition. It seems that the authors are defining a set of probability measures for a sample of $n$ training examples, however I am unsure of what the condition on $\mu$ means in the set definition. Is it basically saying that each Borel set has an equal chance ($\frac{1}{n}$) of manifesting in the training set?
EDIT: My intuition now is that this set definition is basically asserting an i.i.d assumption on the training sample. Is this correct?
 A: Yes, this is basically a statement that the measures are i.i.d.  If we look at some concrete examples we can more clearly see what's going on:


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*${\cal P}_1$ contains all those measures $\mu$ for which $\mu({\cal Z}_1) \geq 1$, i.e. all those measures for which only the first training set is relevant.

*${\cal P}_2$ contains all those measures $\mu$ for which $\mu({\cal Z}_1) \geq \frac{1}{2}$ and $\mu({\cal Z}_2) \geq \frac{1}{2}$ and $\mu({\cal Z}_1\cup {\cal Z}_2) \geq 1$
So far so boring, but when we get to $n=3$:


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*${\cal P}_2$ contains all those measures $\mu$ for which $\mu({\cal Z}_1) \geq \frac{1}{3}$ and $\mu({\cal Z}_2) \geq \frac{1}{3}$ and $\mu({\cal Z}_3) \geq \frac{1}{3}$ and $\mu({\cal Z}_1\cup {\cal Z}_2) \geq \frac{2}{3}$ and $\mu({\cal Z}_1\cup {\cal Z}_3) \geq \frac{2}{3}$ and $\mu({\cal Z}_2\cup {\cal Z}_3) \geq \frac{2}{3}$ and $\mu({\cal Z}_1\cup {\cal Z}_2 \cup {\cal Z}_3) \geq 1$
And now we can clearly see that the aim is to identify those measures that apportion equal weight to the training sets no matter how they're distributed.
