I'm working on the following problem in a book:
Suppose $\mu$ is Lebesgue measure on $[0,1]$. $f_\#\mu$ denotes the pushfroward of lebesgue measure by the function $f$. Find an example of a sequence of functions $f_n:[0,1]\rightarrow [0,1]$ such that $(f_n)_\# \mu = \mu$ but $f_n \rightarrow 1/2$ weak-* in $L^{\infty}[0,1]$. Can $f_n$ have continuous first derivatives?
Since the image measure is lebesgue, the only things I could think that $f_n$ could be are tranlations, rotations, and inversions. Or dialations on a set of measure zero. But I don't know how to enforce the $L^{\infty}$ condition.
Does anyone have any ideas?
EDIT:
by $f_n \rightarrow 1/2$ weak-* in $L^{\infty}([0,1])$, I mean a sequence of functions $\{f_n\} \in L^{\infty}([0,1])$ such that:$$\int_{[0,1]}f_n g \rightarrow \int_{[0,1]}\frac{1}{2}g \quad \forall g\in L^1([0,1])$$
sorry for the delay, I was thinking about your answer. I'm having trouble understanding by $f_n = x$ or $f_n = 1-x$. Certainly these are candidates that statisfy the pushforward condition.
EDIT: Okay the pushforward condition by definition is $f_{\#}\mu[A] = \mu(f^{-1}[A])$ but on the other hand, by ass'n we have: $f_{\#}\mu[A] = \mu[A]$. Thus the condition specifies only that: $\mu(f^{-1}[A]) = \mu[A]$ – i.e. the preimage of $A$ has the same measure as $A$. I think $f(x) = x$ and $f(x) = 1-x$ are the only $C^1$ candidates. But there are other $f$, not $C^1$, that fullfill this criteria consider:
$$
f(x) =
\begin{cases}
x + \frac{1}{2} & 0 \leq x \leq 1/2\\
x – \frac{1}{2} & 1/2 < x \leq 1
\end{cases}
$$
How do I know there isn't some crazy piece-wise function for which the convergence holds?
Best Answer
Observe that another way of stating the pushforward condition is that $f_n$ is a random variable on the measure space $([0,1],\mu)$ whose law is uniform $[0,1]$. Write $X_n=f_n$. Then on the one hand $\mathbb Eg(X_n)=\int_0^1 f_ng$ by definition, and on the other hand since $X_n$ is uniform $[0,1]$ we have that $\mathbb Eg(X_n)=\int_0^1 xg$. Consequently $$ \lim_{n\to\infty}\int_0^1 f_ng=\int_0^1 xg, $$ since it is actually a constant sequence (not depending on the value of $n$) on the left side. But this is obviously not equal to $\tfrac12\int_0^1g$: for example, whenever $g\geq 0$ has positive-measure support contained in $[0,\tfrac14]$ we would have $\int_0^1 xg\leq \tfrac14 \int_0^1 g<\tfrac12 \int_0^1 g.$
Thus there is no such sequence.