Link to image of the Interval
I posted a link to the image if anyone wants to see what the function would look like on a sinlge interval. As the intervals become smaller each triangle will become taller. I hope this is helpful and please feel free to comment if something is wrong.
Let $n \in \mathbb{N}$ and consider the following function f(x) s.t.:
\begin{array}{ll} 0,~x \in \{0,\frac{1}{n+1},~\frac{1}{n} \} \\ 16(n^2+n)^2(x-\frac{1}{n+1}),~~~~ x \in (\frac{1}{n+1},~\frac{1+4n}{4(n^2+n)}) \\ -16(n^2+n)^2(x-\frac{1+4n}{4(n^2+n)})+4(n^2+n),~ x \in [\frac{1+4n}{4(n^2+n)},~\frac{3+4n}{4(n^2+n)}] \\ 16(n^2+n)^2(x-\frac{3+4n}{4(n^2+n)})-4(n^2+n),~ x \in (\frac{3+4n}{4(n^2+n)}~,\frac{1}{n}) \end{array}
Let $A_n=[\frac{1}{n+1},\frac{1}{n}].$ Observe that $\frac{1}{2}(\frac{1}{n}-\frac{1}{n+1})+\frac{1}{n+1}=\frac{1+2n}{2(n^2+n)}$ and is therefore the halfway point of $A_n.$ Likewise the $\frac{1}{4}$ and $\frac{3}{4}$ points on each $A_n$ are $\frac{1+4n}{4(n^2+n)}$ and $\frac{3+4n}{4(n^2+n)}$ respectively. Let $A_n'$ and $A_n''$ be the first and second half of the interval $A_n$ respectively. $f$ on $A_n'$ forms an isosceles triangle above the x-axis with the base on the x-axis and has an area of 1. Likewise $f$ on $A_n''$ forms an isosceles triangle with the same traits except it is below the x-axis. Then $\int_{A_n'}f=\int^{\frac{1+2n}{2(n^2+n)}}_{\frac{1}{n+1}}f=1$ and likewise $\int_{A_n''}f=-1$. Then by additivity over domains of integration we can show that $\int_{A_n'}f+\int_{A_n''}f=\int_{A_n}f=1-1=0~\forall n$. Then by countability of integration we know $\sum\limits^{\infty}_{n=1}\int_{A_n}f= \int_{\cup^{\infty}_{n=1}A_n}f=0$. Thus $\{\int_{[\frac{1}{n},1]}f\} \to 0.$ However $\int_{A_n'}|f|+\int_{A_n''}|f|=\int_{A_n}|f|=1+1=2~\forall n$. Then $\sum\limits^{\infty}_{n=1}\int_{A_n}|f|= \int_{\cup^{\infty}_{n=1}A_n}|f|=\sum\limits^{\infty}_{n=1}2=\infty$. Thus $\int_{[0,1]}|f|=\infty \Rightarrow$ $f$ on $[0,1]$ is not Lebesque integrable.
Let's consider an alternate case in which $\{\int_{[\frac{1}{n},1]}f \}^{\infty}_{n=1}$ converges and $f$ is non negative. Then $\int_{[\frac{1}{n},1]}f \in \mathbb{R}$ and $\int_{[\frac{1}{n},1]}f \le \int_{[\frac{1}{n+1},1]}f~~\forall n \Rightarrow$ $\{\int_{[\frac{1}{n},1]}f \}^{\infty}_{n=1}$ is an increasing sequence of real numbers. Since $[\frac{1}{n},1] \to (0,1]$ and $\{\int_{[\frac{1}{n},1]}f \}^{\infty}_{n=1}$ converges then $\int_{[\frac{1}{n},1]}f \to \int_{(0,1]}f<\infty.$ Since $f$ is non negative then $\int_{(0,1]}f=\int_{(0,1]}|f|=\int_{(0,1]}f^+<\infty$ and $\int_{(0,1]}f^-=0$. Then $f$ is Lebsque integrable. $\square$