I am currently studying for a measure-theoretic probability exam. In an old exam, I encountered the following question:
Let $(U_n)_{n \in \mathbb{N} }$ be i.i.d. sequence of $\text{Uniform}((0,1))$ random variables on a probability space $(\Omega, \mathcal{F}, \mathbb{P})$. Set $\mathcal{F}_n = \sigma(U_0, U_1,...,U_n)$. Set $M_0=U_0$ and $M_n = M_{n-1}^2U_n^{M_{n-1}-1}$. We are asked to show that $(M_n)_{n \in \mathbb{N}}$ is a martingale. Furthermore we are given the hint to use the following theorem:
Let $X,Y \in \mathcal{L}^1(\Omega, \mathcal{F}, \mathbb{P})$, let $\mathcal{G} \subseteq \mathcal{F}$ be a sub-$\sigma$-algebra. Let $X$ be a $\mathcal{G}$-measurable random variable and let the random variable $Y$ be independent of $\mathcal{G}$. Assume that $h ∈ B(\mathbb{R}^2)$ is such that $h(X, Y )\in \mathcal{L}^1(\Omega, \mathcal{F}, \mathbb{P})$. Put $\hat{h}(x) = \mathbb{E} [h(x, Y )]$. Then $\hat{h}$ is a Borel function and $\hat{h}(X)$ is a representative of $\mathbb{E}[h(X, Y )|\mathcal{G}]$.
Adaptedness is clear, but I struggle with the martingale property and the integrability. I have tried the following:
To show the martingale property I think we have to use the theorem given in the hint. Setting $X=M_{n-1}$, $Y=U_n$ and $\mathcal{G}=\mathcal{F}_{n-1}$. Then $M_{n-1}$ is clearly $\mathcal{F}_{n-1}$ measurable and $U_n$ is independent of $\mathcal{F}_{n-1}$. Let $h(x,y)=x^2y^{x-1}$. Then $h$ is measurable because it is a composition of measurable functions. To show the criterium $h(M_{n-1}, U_n )\in \mathcal{L}^1(\Omega, \mathcal{F}, \mathbb{P})$ is the same as integrability of $M_n$, which I have not figured out.
Using the above I get: $$\mathbb{E}[M_n|\mathcal{F}_{n-1}]=\mathbb{E}[M_{n-1}^2U_n^{M_{n-1}-1}|\mathcal{F}_{n-1}]=\mathbb{E}[h(M_{n-1}, U_n )|\mathcal{F}_{n-1}]= \hat{h}(M_{n-1})$$ also, $$\hat{h}(x) = \mathbb{E}[h(x,U_n)]=\mathbb{E}[x^2U_n^{x-1}]=x^2\mathbb{E}[U_n^{x-1}]=x1^x=x$$ Putting this together gives: $\mathbb{E}[M_n|\mathcal{F}_{n-1}]= M_{n-1}$. Where in the last equality, I used that the $x-1$'th moment of a $\text{Uniform}((0,1))$ random variable is given by: $\mathbb{E}[U_n^{x-1}]=\frac{1^x}{x}$. But even then I am not sure how to show integrability. Any help is much appreciated!
EDIT:
I think integrability follows from the above. Note that $M_n \geq 0$, $\mathbb{E}[|M_n|]=\mathbb{E}[M_n]=\mathbb{E}\left[\mathbb{E}[M_n|\mathcal{F}_{n-1}]\right]=\mathbb{E}[M_{n-1}]=\mathbb{E}[M_{0}]=\frac{1}{2} < \infty$.
As pointed out by Jose Avilez, the problem in your proof of integrability of $M_n$ rests on the fact that $\mathbb E\left[M_n\mid\mathcal F_{n-1}\right]=M_{n-1}$, which uses the mentioned theorem, which requires integrability of $M_n$ hence the reasoning would be circular.
The difficulty here is that the exponent $M_{n-1}-1$ can be negative, like for $n=1$, $M_1=U_0^2 U_1^{U_0-1}$. To overcome this problem, fix $n$ and define
$$ a_k:=\mathbb E\left[M_{n-1}^2 U_n^{M_{n-1}-1}\mathbf{1}\left\{2^{k-1}\leqslant M_{n-1}<2^{k} \right\}\right], k\in\mathbb Z. $$ Note that $a_k$ is finite because $$ M_{n-1}^2 U_n^{M_{n-1}-1}\mathbf{1}\left\{2^{k-1}\leqslant M_{n-1}<2^{k} \right\}\leqslant 2^{2k}U_n^{2^{k-1}-1}\mathbf{1}\left\{2^{k-1}\leqslant M_{n-1}<2^{k} \right\}, $$ which gives; by independence between $U_n$ and $M_{n-1}$, that $$ a_k\leqslant 2^{2k}\mathbb{E}\left[U_n^{2^{k-1}-1}\right]\mathbb P\left\{2^{k-1}\leqslant M_{n-1}<2^{k} \right\} $$ hence for $k\leqslant 0$, $$ a_k\leqslant 2^{k+1}\mathbb P\left\{2^{k-1}\leqslant M_{n-1}<2^{k} \right\}\leqslant 2\mathbb P\left\{2^{k-1}\leqslant M_{n-1}<2^{k} \right\} $$ and the series $\sum_{k\leqslant 0}a_k$ is convergent. For the series $\sum_{k>0}a_k$, simply note that this time, the exponent $M_{n-1}-1$ is non-negative hence $a_k\leqslant \mathbb P\left\{2^{k-1}\leqslant M_{n-1}<2^{k} \right\}$.