I'm wondering about a perhaps trivial question.
Suppose $f(x) = 1$ when $x \geq 0$ and $f(x) = -1$ when $x < 0$. Since
$$\int_{\mathbb{R}}|f(x)|dx = \infty $$
Clearly $f$ is not integrable on $\mathbb{R}$ w.r.t. Lebesgue measure.
Now we define $I_{n}(x)$ to be the characteristic function of $[-n,n]$, i.e. $I_n(x)$ equals to $1$ iff $x \in [-n,n]$ and $I_n(x) = 0$ everywhere else. Denote $f_+(x) = \max(f(x),0)$ and $f_-(x) = \min(f(x),0)$, then $f(x) = f_-(x)+f_+(x)$, and we have $f_+(x)I_n(x)$ converges to $f_+(x)$ everywhere monotonically, and the similar relationship holds for the negative part.
Now it is the part I'm confused. By Levy's monotone convergence theorem, we obtain $$\lim_{n\rightarrow\infty}\int_{0}^{n} f_+(x)dx =\lim_{n\rightarrow\infty}\int_{\mathbb{R}} f_+(x)I_n(x)dx = \int_{\mathbb{R}}f_+(x)dx$$ and similarly, $$\lim_{n\rightarrow\infty}\int_{-n}^{0} f_-(x)dx =\lim_{n\rightarrow\infty}\int_{\mathbb{R}} f_-(x)I_n(x)dx = \int_{\mathbb{R}}f_-(x)dx$$ And therefore, $$\int_{\mathbb{R}}f(x)dx = \int_{\mathbb{R}}f_+(x)dx+\int_{\mathbb{R}}f_-(x)dx=\lim_{n\rightarrow\infty}\int_{0}^{n} f_+(x)dx +\lim_{n\rightarrow\infty}\int_{-n}^0 f_-(x)dx = \lim_{n\rightarrow\infty}\int_{-n}^{n} f(x)dx = 0$$
Therefore $f$ is integrable, which makes a contradiction.
Can anyone tell me what's wrong with the whole process?
Also, is $f$ Riemann integrable, or integrable under any other sense? It seems pretty natural to think that the integral of $f$ equals to zero, since $f(x) = -f(-x)$ everywhere. However it seems Lebesgue integral is not convenient to handle such functions.
Somewhere down the line you wrote $\lim_{n\to\infty}\int_0^n f_+(x)\,dx$ and manipulated that as if it was a number. Would you have done that if you had written it as $\lim_{n\to\infty}n$? You can rewrite your argument as $$\infty-\infty=\lim_{n\to\infty}n+\lim_{n\to\infty}(-n)= \lim_{n\to\infty}0=0.$$