This question Sum of Countable Increasing Absolutely Continuous Functions is Absolutely Continuous says if the functions are all AC, then the sum is also AC. Then if the answer is not true for my question, we only need to consider the case of continuous singular functions. However, I have no clue how to deal with them since I only know the Contor function as an example of continuous singular function, and the summation of the varients of it seems cannot lead to a contradiction.
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If $\sum_n f_n$ converges pointwise, then for every closed interval $[a,b]$, we have for each $\epsilon>0$ that for some $N$, and each $m,n$ with $N<m\leq n$, that $$\left|\sum_{k=m}^nf_k(x)\right|\leq \epsilon,\tag{1}$$
for $x=a$ and $x=b$.
Then, by monotonicity, we have for each $x\in [a,b]$,
$$ -\epsilon\leq \sum_{k=m}^nf_k(a) \leq \sum_{k=m}^nf_k(x) \leq \sum_{k=m}^nf_k(b)\leq \epsilon,$$ and so (1) holds for all $x\in [a,b]$. Therefore the sum converges uniformly on compact sets, and so the limit is continuous.
M W
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nice observation! I am also wondering whether this can be generalized to the continuous case, i.e. for any fixed $0\le y \le 1$, a positive function $f(x,y)$ is increasing and continuous w.r.t. $x$, then does $\int_0^1 f(x,y) dy$ necessarily be continuous in $x$? – qdmj Nov 05 '23 at 02:22
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@qdmj Yes, and you don’t need positivity for $f$ - as long as for each $x$ your function is integrable wrt $y$ over $[0,1]$ that should already be enough. This follows from the dominated convergence theorem (all the functions $f_x(y):=f(x,y)$ between $x=a$ and $x=b$ are dominated by $|f_a|+|f_b|$) – M W Nov 05 '23 at 17:35