The continuity of the function on $\mathbb{R}$: $I(y)=\displaystyle \int \limits_0^1sgn(x-y)\,dx $ Investigate the continuity of the function on $\mathbb{R}$:
$$I(y)=\displaystyle \int \limits_0^1\text{sgn}(x-y)\,dx $$
I have some ideas:

*

*Considering $[a,b], \, \forall b>a>1$. Then, $\text{sgn}(x-y)=-1, \forall (x,y) \in [0;1]\times[a;b]$.

$\qquad$So, $I(y)$ is continuous on $[a,b]$, or $I(y)$ is continuous at all point $y>1$.


*Prove similar, $I(y)$ is continuous at all point $y<0$.

*On $[0;1]$, I don't know how to prove.

 A: $$
I(y) = \left\{
\begin{array}{rl}
1, & y \leq 0 \\
\int_0^y (-1)dx + \int_y^1 dx = 1-2y, & 0 < y \leq 1 \\
-1, & y > 1 
\end{array}
\right.
$$
This function has 2 points of possible discontinuity: $y = 0$ and $y = 1$.
$$
\lim\limits_{y \rightarrow -0}I(y) = 1 = \lim\limits_{y \rightarrow +0}I(y).
$$
$$
\lim\limits_{y \rightarrow 1-0}I(y) = -1 = \lim\limits_{y \rightarrow 1+0}I(y).
$$
This means that $I(y)$ is continuous.
A: 
I know the answer I am going to give looks a sledgehammer for killing a fly.
My intention here is to show that there exists higher level concepts/tools/properties that, even if you have not already met them can radically change the point of view. Anyhow, it is tools that you will meet in the coming year(s) if you study Calculus at a higher level, in particular in connection with Fourier/Laplace transforms.
The form of the integral (and as well the graphical representation) call for using convolution.
Indeed, your integral can be written under the form:
$$I(y)=\displaystyle \int \limits_{-\infty}^{\infty}Rect(x)\text{sgn}(x-y)\,dx=-\displaystyle \int \limits_{-\infty}^{\infty}Rect(x)\text{sgn}(y-x)\,dx$$
where $Rect$ is the characteristic function of interval $[0,1]$
($Rect(x):=1$ if $x \in [0,1]$, $0$ otherwise).
With appropriate notations:
$$I=- Rect \star sgn$$
Such a convolution is known to be continuous but let us prove it by using (once more) new tools/properties. The derivative of a convolution product can be obtained by convolving with one of the derivatives of its components (the other one remaining unchanged):
$$(f \star g)'=f \star g' \ \text{giving here} \ I'=- Rect \star sgn'=- Rect \star 2 \delta$$
Where the $2 \delta$ factor accounts for the sudden jump of intensity $+2$ at $x=0$.
As Dirac $\delta$ in neutral for convolution:
$$I'=-2 Rect$$
a property that can help you to build the curve (in a revese way, you can verify that the derivative is $-2$ on [0,1] and zero elsewhere).
In fact, we have to take an integration constant equal to $1$ because the value $I(0)=1$ can be computed dierectly on the integral definition.
A: Probably easier under the form
$$I(y)=\int \limits_{-y}^{1-y}\text{sgn}(x)\,dx.$$
The antiderivative of the sign function is continuous, as is the antiderivative of any piecewise continuous function. And the difference of two continuous functions is continuous.

A: Another possible approach: For $y_1 < y_2$ set $a = \min(y_1, 0)$ and $b = \max(y_2, 1)$. Then
$$
 0 \le I(y_1) - I(y_2) = \int_0^1 (\underbrace{\operatorname{sgn}(x-y_1)-\operatorname{sgn}(x-y_2)}_{\ge 0}) \, dx \\
\le \int_a^b (\operatorname{sgn}(x-y_1)-\operatorname{sgn}(x-y_2)) \, dx
= \int_{y_1}^{y_2} 2 \, dx = 2(y_2-y_1) \, .
$$
A: It should easy enough to obtain an explicit form for $I(y)$ by considering all $y \in \mathbb R$ and three cases:

*

*$y < 0$ which implies $x-y > 0$;

*$y > 1$ which implies $x-y < 0$;

*$0 \leqslant y \leqslant 1$, and in this case divide the integral into $[0,y]$ and $[y,1]$.

Then just examine $I(y)$ when $y=0$ and $y=1$.
