Show by definition that $f:[0,1]\to\mathbb R$ is measurable 
Show by definition  that the function $f:[0,1]\to \mathbb R$ defined by $$f(x)=\begin{cases}\frac{1}{x} & 0<x<1\\3 & x=0\\5 & x=1\end{cases}$$is measurable.

Let  $\alpha$ be any real number. Then 
$$\{f>\alpha\}=\begin{cases} [0,1] & \alpha<1 \\ [0,1/\alpha] &\alpha\ge 1 \end{cases}$$
Please check my calculation and detect if something wrong there. Also comment if there are no mistake.

Your calculations are still not correct ... since this exercise is not that difficult, this makes me believe that you don't really know what you are doing.
In order to show measurability of $f$, we have to show that the preimages
$$\{f>\alpha\} := \{x \in [0,1]; f(x)>\alpha\}$$
are Borel sets for any $\alpha \in \mathbb{R}$. To get some first intuition how these sets look like, it is a good idea to draw a picture.

For each fixed $\alpha$ we have to find the points $x \in [0,1]$ such that the value $f(x)$ is (stricly) above the green line; above you see pictures for $\alpha=2$ and $\alpha=4$.


*

*Case 1: $\alpha<1$. Since $f(x) \geq 1>\alpha$ for any $x \in [0,1]$ we have $$\{f>\alpha\} = [0,1]$$ for any $\alpha<1$, this agrees with your calculation.

*Case 2: $\alpha \in [1,3)$. Using the monotonicity of $f$ on $(0,1)$ and the fact that $f(0)=3 >\alpha$ and $f(1)=5>\alpha$, we find $$\{f>\alpha\} = (0,\alpha^{-1}) \cup \{0\} \cup \{1\}$$ for $\alpha \in [1,3)$.

*Case 3: $\alpha \in [3,5)$. Note that $f(0) = 3 < \alpha$ and therefore $0 \notin \{f>\alpha\}$. Again the monotonicity on $(0,1)$ and the fact that $f(0)=5>\alpha$ yields $$\{f>\alpha\} = (0,\alpha^{-1}) \cup \{1\}.$$

*Case 4: $\alpha \geq 5$. For any such $\alpha$, we have $\{f>\alpha\} = (0,\alpha^{-1})$.


Since $(0,\alpha^{-1})$, $\{0\}$, $\{1\}$ are Borel sets, this shows that $f$ is (Borel)measurable.
