First off! This is a homework question, so I DO NOT want an answer to the question I'm writing, I really just want an explanation of the final bit (which I'll make clear).
So if we have $T:[0,1)\to [0,1),Tx=2x \text{mod} 1$ which preserves Lebesgue measure $m$, then we have $\phi:[0,1)\to[0,1)$ defined as $\phi(x)=x^3$. Find $$\lim_{n\to ∞} \frac{1}{n} \sum_{j=0}^{n-1} \phi(T^j x)$$ for $m$ almost every point $x\in [0,1)$.
So essentially, my question is "what does it mean 'Find ** for m almost every point x'?" My question really is that I don't know what I'm computing? Does it mean compute $m(***)?$ If that's the case, I think I have a few ideas....
In measure theory, we say a property holds "almost everywhere" if the set of elements for which it doesn't hold has measure zero. Clearly, this definition depends on the measure, so we might say "$m$-almost everywhere" if it's not clear from context.
In this case, what we're saying is, find an expression $f(x)$ such that the equation \[f(x) = \lim_{n\to ∞} \frac{1}{n} \sum_{j=0}^{n-1} \phi(T^j x) \tag{1}\] holds $m$-almost everywhere in $[0,1)$, i.e. $m(\{x \in [0,1) : \text{(1) doesn't hold}\}) = 0$.
Of course such an $f$ will be non-unique. But that's probably okay – whoever is asking the question doesn't mind which you choose.