Can you get a discontinuous function like the square wave in Fourier analysis, if all the functions are decreasing or monotone? For each $\ k\in\mathbb{N},\ $ let $\ f_k:(0,1)\to\mathbb{R}\ $ be a continuous function and also, for each $\ x\in (0,1),\ \displaystyle\sum_{k\in\mathbb{N}} f_k(x)\ $ converges. Then from Fourier analysis, we know that $\ g(x):=\displaystyle\sum_{k\in\mathbb{N}} f_k(x)\ $ need not be continuous: for example, $\ g(x)\ $ could be a square wave. This prompts me to ask the following question:

For each $\ k\in\mathbb{N},\ $ let $\ f_k:(0,1)\to\mathbb{R}\ $
be a continuous, decreasing function and also, for each $\ x\in
 (0,1),\ \displaystyle\sum_{k\in\mathbb{N}} f_k(x)\ $ converges. Then
$\ g(x):=\displaystyle\sum_{k\in\mathbb{N}} f_k(x)\ $ is continuous.

Is this true? Further: what if we replace decreasing with monotone?
My further question - if it's true - implies that non-monotonicity is required for $\ g(x)\ $ to be discontinuous.
 A: I think it is true: Fix $x_0 \in (0,1)$ and choose $0<a<x_0<b<1$, so
$x_0 \in (a,b) \subseteq [a,b] \subseteq (0,1)$. For $x \in [a,b]$ and each $k$ we have $f_k(x) \ge f_k(b)$. The series $\sum_{k=1}^\infty f_k(b)$ and  $\sum_{k=1}^\infty (f_k(a)-f_k(b))$ are both convergent, and since
$$
0 \le f_k(x) -f_k(b) \le f_k(a)-f_k(b) \quad (x \in [a,b])
$$
the Weierstraß M-test shows that $\sum_{k=1}^\infty (f_k(x)-f_k(b))$ is uniformly convergent on $[a,b]$. Hence
$$
x \mapsto g(x):=\sum_{k=1}^\infty f_k(x) = \sum_{k=1}^\infty (f_k(x)-f_k(b)) + \sum_{k=1}^\infty f_k(b)
$$
is continuous in $x_0$. Since $x_0 \in (0,1)$ was arbitrary we have that $x \mapsto g(x)$ is continuous on $(0,1)$.
Edit: I think I have a counterexample for the mixed case (all $f_k$ monotone): Consider the interval $(0,2\pi)$ (instead of $(0,1)$) and consider the Fourier series
$$
h(x):=\sum_{n=1}^\infty (-1)^{n+1} \frac{\sin(nx)}{n} \quad (x \in(0,2\pi)),
$$
which is known to be pointwise convergent and $h$ is discontinuous in $\pi$. For each $n$ the functions
$$
x\mapsto u_n(x):=\frac{1}{2}(\frac{\sin(nx)}{n}+x), \quad x\mapsto v_n(x):=\frac{1}{2}(\frac{\sin(nx)}{n}-x)
$$
are monotone (increasing and decreasing, respectively). Now construct a series $\sum_{k=1}^\infty f_k$ the following way:
$$
u_1+v_1-\frac{1}{2}u_2-\frac{1}{2}v_2-\frac{1}{2}u_2-\frac{1}{2}v_2 + 
\frac{1}{3}u_3+\frac{1}{3}v_3+\frac{1}{3}u_3+\frac{1}{3}v_3+\frac{1}{3}u_3+\frac{1}{3}v_3 - \dots
$$
$$
+ (-1)^{n+1}\frac{1}{n}u_n+ (-1)^{n+1}\frac{1}{n}v_n+ \dots +  (-1)^{n+1}\frac{1}{n}u_n+(-1)^{n+1}\frac{1}{n}v_n + \dots,
$$
that is $f_1=u_1$, $f_2=v_1$, $f_3=-u_2/2$, $\dots$
Now a partial sum of $\sum_{k=1}^\infty f_k$ differs from a suitable
partial sum of $\sum_{n=1}^\infty (-1)^{n+1}(u_n+v_n)$ by an expression of the form
$$
(-1)^{n+1}\frac{l}{n}\frac{\sin(nx)}{n}, \quad l \in \{0, \dots,n-1\}
$$
or
$$
(-1)^{n+1}\left(\frac{l}{2n}\frac{\sin(nx)}{n} + \frac{x}{2n}\right), \quad l \in \{0, \dots, 2n-1\}.
$$
Hence $\sum_{k=1}^\infty f_k(x)$ is pointwise convergent with limit $h(x)$.
