integration - function which is Riemann integrable

Consider $f:[-1,1]\to\mathbb{R}$, $x\mapsto \begin{cases} 1, & \text{if } x=0 \\ 0 & \text{else } \end{cases}$ I want to know why f is Riemann integrable and I tried something. First of all we had the definition in lecture: $\int_{-1}^{1^*} f(x)\,dx=inf\{\int_{-1}^1 \phi(x)\,dx;\phi:[-1,1]\to\mathbb{R} \;\text{step function}, f\le\phi \}$ and $\int_{-1^*}^{1} f(x)\,dx=inf\{\int_{-1}^1 \phi(x)\,dx;\phi:[-1,1]\to\mathbb{R} \;\text{step function}, \phi\le f \}$. Now the bounded function f is Riemann integrable, if $\int_{-1}^{1^*} f(x)\,dx=\int_{-1^*}^{1} f(x)\,dx$.

My first try: consider the step function $\phi:[-1,1]\setminus\{0\}\to\mathbb{R},\; x\mapsto 0$. It is $f=\phi$ everywhere on $[-1,1]\setminus \{0\}$. Therefore it is
$\int_{-1}^{1^*} f(x)\,dx=inf\{\int_{-1}^1 \phi(x)\,dx;\phi:[-1,1]\to\mathbb{R} \;\text{step function}, f\le\phi \}=\int_{-1}^{1} \phi(x)\,dx=0$ and $\int_{-1^*}^{1} f(x)\,dx=\int_{-1}^{1} \phi(x)\,dx=0$. Therefore f is Riemann-integrable and $\int_{-1}^{1} f(x)\,dx=0$. Could you help me, to correct my solution?

I'm not sure if it is ok, because my step function isn't defined in $x=0$. But we say that $\phi:[-1,1]\to\mathbb{R}$ is a step function if you have a partition of the interval $[-1,1]$, $x_0=-1Maybe I have to define $\phi$ in $x=0$ too but take $x=0$ as one of the $x_i's$.

I will try next to do this with Riemann Sums. Regards