Solution

First we define the lists for the definition set and the value set, with the requirements, that the value set is distributed symmetrically around $x=0$ and that the integral over the function is unity.

NN = 3; (*number of intervals per quadrant*)
XL = N[Range[-1, 0, 1/NN]];
XL = XL[[1 ;; Length[XL] - 1]];
XR = N[Range[0, 1, 1/NN]];
XR = XR[[2 ;; Length[XR]]];
X = Join [XL, XR]];
YL = RandomReal[1/(NN), NN];
YR = Reverse [YL];
Y = Join [YL, YR];
HalfStep = N[1/(2*NN)];
RhoArr = Transpose[{X, Y}];

Now we define the desired function

g[x_] := Piecewise[
Table[{Y[[k]], 
Which[k == 1, X[[k]] <= x <= X[[k]] + HalfStep, 
k == Length[X], X[[k]] - HalfStep <= x <= X[[k]], 
True, X[[k]] - HalfStep <= x <= X[[k]] + HalfStep]}, 
{k, 1, Length@X}]];

and furthermore

Zer = (1 - NIntegrate[g[x], {x, -1, 1}])/(2*HalfStep)(*For the integral to be 1*);
X = Insert[X, 0, NN + 1]; 
Y = Insert[Y, Zer, NN + 1];
RhoArr = Transpose[{X, Y}];

NIntegrate[g[x], {x, -1, 1}]]
Plot[g[x], {x, -1, 1}, PlotRange -> {0, 0.5}]

So everything works.

Thanks to all involved.

Old response - Question 2

First: I am the one who asked the question. Unfortunately my guest account, with which I wrote this post, and the current real account, which is the one from the guest account, are not connected, although the email matches and I just added a password to the guest account. Therefore I can neither edit my question nor write a comment or ask another question about it. I am sorry for this mess, it is the first time ever that I am in a forum. If anyone can help me clean up this mess, this would be helpful.

To answer m_Goldberg's question above and provide some clarity: Here is the code where the $X$ and $Y$ lists are defined, which in turn define the 2-d list RhoArr; the definitions aim to create a 2-d list with the contents

{definition set, value set}

where the value set is to be distributed symmetrically around $x=0$.

NN = 3;
XL = N[Range[-1, 0, 1/NN]];
XL = XL[[1 ;; Length[XL] - 1]];
XR = N[Range[0, 1, 1/NN]];
XR = XR[[2 ;; Length[XR]]];
X = Join [XL, XR]];
YL = RandomReal[1/(2*NN), NN];
YR = Reverse[YL];
Y = Join[YL, YR];
Zer = 1 - Total [Y];
X = Insert[X, 0, NN + 1];
Y = Insert[Y, Zer, NN + 1];
RhoArr = Transpose[{X, Y}]
HalfStep = N[1/(2*NN)];

The problem is actually simple, at least in other programming languages, I just don't know how to solve it in Mathematica.

A short and precise summary of the problem: I have a 2-d list (RhoArr) and want to define a step function from it, which has an input value $x$ (see the mathematically written function $f(x)$ in the original post), which then can be plotted via $x$ and integrated via $x$. My above solution using the functions $f(x)$ and $g(x)$ is obviously bad, because it can only be plotted, but not integrated. The plot looks like this

However, if you integrate this function, the error messages from the original question will appear.

In general: such a mess won't happen to me again and I apologize for that! I will also try to formulate my problems more clearly and completely. All that remains is my justification that I have never done something like this before.

Here's a simple way: use an interpolating function (with order 0).

rhoArr = {{-1, 0.3}, {-0.5, 0.2}, {0, 0.1}, {0.5, 0.2}, {1, 0.3}}; 
f = Interpolation[rhoArr, InterpolationOrder -> 0];
Plot[f[x], {x, -2, 2}]

You can use the function f just like any other function. In particular you can integrate it. Here is the integral from -1 to 1:

Integrate[f[x], {x, -1, 1}]
0.4