show that $$ \sum_{n=1}^{\infty} \chi_{\left[n, n+1 / n^{2}\right]}(x) $$ is pointwise converge for $x \in \mathbb{R}$ while $\chi_{\left[n, n+1 / n^{2}\right]}(x)$ is the indicator function
The solution in the book :
$$ \chi_{\left[n, n+1 / n^{2}\right]}(x)=\left\{\begin{array}{ll} 1 & n \leq x \leq n+\frac{1}{n^{2}} \space \\ 0 & \text { otherwise } \end{array}\right. $$ thus for every $x$: $$ \sum_{n=1}^{N} \chi_{\left[n, n+1 / n^{2}\right]}(x)=\sum_{n=1}^{N_{0}} \chi_{\left[n, n+1 / n^{2}\right]}(x) $$ for $ N \geq N_{0} \geq x+1 $ and thus the series is pointwise.
I dont understand how:
$\sum_{n=1}^{N} \chi_{\left[n, n+1 / n^{2}\right]}(x)=\sum_{n=1}^{N_{0}} \chi_{\left[n, n+1 / n^{2}\right]}(x)$ imply that we have pointwise converge
what the inequality $ N \geq N_{0} \geq x+1 $ mean to us in this problem, why is there $+1$ factor?
To prove pointwise convergence of your series, I'd look at what happens at different points $x$. The "worst" case is when $x=2$ and your series is $1+1+0+0+0+0+0+\dots$, which converges to $2$. For all other values of $x$, the series is either all zeros (converging to $0$) or a single $1$ and all the rest zeros (converging to $1$). So the series converges for every value of $x$, and that's what pointwise convergence means.
The solution you quoted uses the fact that, if the terms of a series are all zeros from some point on, then, no matter what happened before that point, the series will converge because the partial sums are equal from that point on. The quoted solution just writes that observation out in impressive notation.