Determine if the function $f(x)$ is integrable in $[0,1]$ Determine if the function $f(x)$ is integrable in $[0,1]$
$$\displaystyle f( x) =\begin{cases}
\frac{1}{x^{2}} & x >0\\
0 & x=0
\end{cases}$$

Attempt:
Let's check the limits of each side $0^+$ to show that $\frac{1}{x^2}$ is not bounded.
$$\displaystyle \lim _{x\rightarrow 0^{+}}\frac{1}{x^{2}} =\infty $$
Therefore, the function is not bounded when $\lim_{x\rightarrow 0^{+}}$
, and is not integrable in $[0,1]$
In addition, there is a well-known theorem:

If $f$ is a continuous function except for the finite number of unconsciously points $\Rightarrow$ $f$ is integrable.

Which contradicts the attempt.

The theorem that you have in mind states that a bounded function from an interval $[a,b]$ into $\Bbb R$ is Riemann-integrable if and only if its set of points of discontinuity has Lebesgue measure $0$. So, there is no contradiction here. Besides, note that the concept of Riemann-integrable function is defined only for bounded functions anyway. My guess is that you had improper integrals in mind.
Besides, the assertion $\lim_{x\to0^-}\frac1{x^2}=\infty$ is irrelevant here, since the domain of $f$ is $[0,1]$.
