= Solution
Assume $\mathbb E|g(X)|<\infty$, or use nonnegative $g$ with extended expectations. Let $p_{ij}=\mathbb P(X=x_i,Y=y_j)$ and $p_j=\mathbb P(Y=y_j)$. For $p_j>0$, the discrete <conditional expectation> is $\mathbb E[g(X)\mid Y=y_j]=\sum_i g(x_i)p_{ij}/p_j$. Therefore <Fubini's theorem>, or <Tonelli theorem> in the nonnegative case, gives the <tower property of conditional expectation>:
$$
\boxed{\mathbb E[\mathbb E[g(X)\mid Y]]=\sum_{j:p_j>0}\sum_i g(x_i)p_{ij}=\sum_i g(x_i)\mathbb P(X=x_i)=\mathbb E[g(X)].}
$$
Values of a <conditional expectation> on zero-probability conditioning events do not affect the sum. An unrestricted signed $g$ whose expectation is undefined is outside this assertion.
Write $M(x)=\mathbb E[N(x)]$. The waiting time for two increments larger than $1/2$ dominates $N(x)$ for $0\leq x\leq1$ and has mean $4$, so $M$ is finite. Conditioning on the first increment, one step is always consumed; if that increment is below $x$, the remaining independent sequence has the original law. Thus <first-step analysis> and the <tower property of conditional expectation> give
$$
M(x)=1+\int_0^xM(x-u)\,du=1+\int_0^xM(s)\,ds.
$$
This first makes $M$ continuous and then gives $M'=M$, with $M(0)=1$. Hence the <uniform-sum crossing time below one> has
$$
\boxed{\mathbb E[N(x)]=e^x,\qquad 0\leq x\leq1.}
$$
An independent explanation uses the <tail-sum formula>: $\mathbb P(N(x)>n)=\mathbb P(U_1+\cdots+U_n\leq x)=x^n/n!$ for $x\leq1$, since this region is an $n$-dimensional simplex inside the unit cube. Summing from $n=0$ gives $e^x$. In particular the strict-crossing convention gives $N(0)=1$ almost surely and $\mathbb E N(1)=e$.
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