Let $f_j(x) =\left\{x + \frac{1}{2} \left\{x + \cdots + \frac{1}{j}\left\{x + \frac{1}{j+1}\right\}\cdots\right\}\right\}$ for $j > 0$. Define
$$g_0 = 0; \quad g_j(x) = \left\{x + \frac{1}{2} \left\{x + \cdots + \frac{1}{j}\left\{x\right\}\cdots\right\}\right\} \text{ for } j > 0.$$
Finally, define the sequence
$$
r_k = \frac{1}{(k+1)!\sum_{n=k+1}^{\infty} \frac{1}{n!}}
$$
for $k \geq 0$, and write $\alpha = e - 1$ for convenience. Note $r_k$ is increasing and that $\lim_{k \to \infty} r_k = 1$.
Lemma. $f_j \to \alpha(x - r_m) + g_n(r_m)$ pointwise on $(r_m, r_{m+1})$.
Proof. Denote the domain of continuity of a function $a \colon \mathbb{R} \to \mathbb{R}$ by
$$\mathcal{C}(a) = \{x \in [0,1] \mid a \text{ is continuous at } x
\}.$$
Note that
- $\mathcal{C}(a \circ b) = \mathcal{C}(b) \cap b^{-1}[\mathcal{C}(a)]$ for $a,b \colon \mathbb{R} \to \mathbb{R}$, and
- $\mathcal{C}(a + b) = \mathcal{C}(a)$, if $b$ is continuous.
Of interest to us is the fact that $\mathcal{C}(x \mapsto \{x\}) = \mathbb{R} - \mathbb{Z}$. Now, given a function $a \colon \mathbb{R} \to \mathbb{R}$, define the $\lambda$-crease of $a$ to be the map $T_\lambda[a]$ taking $x \mapsto \frac{1}{\lambda}\{x + a(x)\}$. Then we may compute
$$\mathcal{C}(T_{\lambda}[a]) = \mathcal{C}(a) \cap \{x \mid x + a(x) \not\in \mathbb{Z}\}.$$
Observe that $f_j$ is actually the result of performing multiple creases on $a \equiv \frac{1}{j+1}$. Specifically, $f_j = T_1 T_{2} \cdots T_{j} \left[\frac{1}{j+1}\right]$. Then it follows that
$$\mathcal{C}(f_j) = \left\{ x \bigm| x + T_{k} \cdots T_{j}\left[\tfrac{1}{j+1}\right] < 1 \text{ for all } 1 \leq k \leq j \right\}.$$
Now, we can determine all the points of discontinuity from solutions to the equations
$$f_{j, k}(x) \overset{\text{def}}{=} x + T_{k} \cdots T_{j}\left[\tfrac{1}{j+1}\right] = 1; \quad f_{j, j+1} \overset{\text{def}}{=} x + \tfrac{1}{j+1} = 1.$$
I claim that $f_{j,k}(x) < 1$ for all $x \in (0, r_{k-3})$. To see this, write $f_{j,k}(x) = x + \frac{1}{k}\{f_{j, k-1}(x)\}$, and note that for $x < r_{k-3}$, we have
$$f_{j,k}(x) < r_{k-3} + \tfrac{1}{k} < 1,$$
since
$$\frac{1}{r_{k-3}} = \frac{1}{k-1}\left(k + \frac{1}{k} + \frac{1}{k(k+1)} + \frac{1}{k(k+1)(k+2)}+\cdots\right) > \frac{k}{k-1}.$$
As $(r_k)_{k \in \mathbb{N}}$ is increasing, it follows that for any integer $k' \in \{k, \ldots, j\}$, one also has that $f_{j, k'}(x) < 1$ for all $x \in (0, r_{k-3})$. Then if $m(k, x)$ is the greatest integer such that $f_{j, m}(x) = 1$, we must have $m(k,x) < k \leq j$ whenever $x \in (0, r_{k-3})$. In that case, we have
\begin{align*}
f_{j,j+1}(x) &= x + \tfrac{1}{j+1} \\
f_{j,j}(x) &= x + \tfrac{1}{j}\{f_{j, j}(x)\} = x + \tfrac{1}{j}\left( x + \tfrac{1}{j+1} \right) \\
&{\hspace{6pt}\vdots} \\
f_{j, m}(x) &= x + \tfrac{1}{m}\left(x + \tfrac{1}{m+1}\left(x + \cdots + \tfrac{1}{j} \left(x + \tfrac{1}{j+1}\right) \cdots \right) \right) = 1.
\end{align*}
This is a linear equation with a unique solution $s_{j, m}$, which may or may not be in the desired interval. Therefore, there are at most $k$ points of discontinuity of $f_j$ in the interval $(0, r_{k-3})$. However, by explicitly calculating these solutions for large $j$, we can determine that exactly $k - 2$ points of discontinuity lie in this interval. In particular, we have that for $m < k$,
$$s_{j,m} = \frac{\frac{1}{(m-1)!} - \frac{1}{(j+1)!}}{\frac{1}{(m-1)!} + \frac{1}{m!} + \cdots + \frac{1}{j!}} \to r_{m-2},$$
and moreover, that $s_{j,m}$ is increasing in $j$, since $\frac{c}{d} > \frac{a}{b}$ implies $\frac{a+c}{b+d} > \frac{a}{b}$ and
$$\frac{\frac{1}{(j+1)!} - \frac{1}{(j+2)!}}{\frac{1}{(j+1)!}} = 1 - \frac{1}{j+2} > 1-\frac{1}{m+3} > s_{j,m}.$$
Therefore, for large $j$, the only points of discontinuity of $f_j$ in the interval $(0, r_{k-3})$ are the roots $s_{j,2}, \ldots, s_{j,k-1}$. By a similar argument, we may determine the points of discontinuity of $g_j$, and realize that they do not include any $r_{k}$. This can be done through wrangling several inequalities, but the quickest way to see this is that the points of discontinuity of $g_j$ will all be rational, whereas each $r_k$ is irrational.
We can now compute the pointwise limit of $f_j$. Suppose that $x_0 \in (r_m, r_{m+1})$ Then there exists some integer $J \geq m + 3$ such that for all $j \geq J$, we have $x_0 \in (s_{j, m+2}, s_{j, m+3})$. Since $f_j$ is right-continuous and linear on $[s_{j, m+2}, s_{j, m+3})$, we may write
$$f_j(x) = \alpha_j (x - s_{j, m+2}) + f_j(s_{j, m+2}) = \alpha_j (x - s_{j, m+2}) + g_{m}(s_{j, m+2}),$$
where $\alpha_j = \sum_{n=1}^{j} \frac{1}{n!}$ is the slope of $f_j$. Since $g_m$ is continuous at $r_m$, we may take the limit to obtain
$$\lim_{j \to \infty} f_j(x) = \alpha (x - r_m) + g_m(r_m),$$
as desired.
The bounded convergence theorem implies that the limit of $I_j = \int_{0}^{1} f_j(x) \, \mathrm{d}x$ exists, and is equal to $\int_{0}^{1} f(x) \, \mathrm{d}x$. By the previous lemma, we may integrate $f$ over each interval $(r_m, r_{m+1})$, yielding
$$
\lim I_j = \int_{0}^{1} f(x) \, \mathrm{d}x = \frac{1}{2\alpha} + \sum_{n=0}^{\infty} \left[g_n(r_{n})\left(r_{n+1} - r_n\right) + \frac{\alpha}{2}(r_{n+1}-r_n)^2\right].
$$
