Step Functions — Question 6

Question 6 – Solution

Strategy. Separate equality at every point from equality except at isolated points. An ordinary integral cannot recover the latter point values.

Step 1: Locate the discrepancy. For $t\ne 1$, $f(t)=g(t)$. At one, $g(1)=1$ but $f(1)=7$. Both functions have left limit zero and right limit one. Thus $g$ is right-continuous, whereas the chosen value of $f$ agrees with neither limit.

Step 2: Transform the interval contribution. A single changed value contributes nothing to an integral. Substituting $v=t-1$ in the tail gives, for both functions, $$\boxed{F(s)=G(s)=\frac{e^{-s}}{s+1},\qquad s>-1.}$$ At $s=-1$, the weighted tail is the positive constant $e$; below $-1$ it grows exponentially. Thus the stated convergence interval is exact. The isolated value seven cannot change this conclusion.

Step 3: Give a pointwise correction and prove the limitation. If $\mathbf 1_{\{1\}}$ is one only at $t=1$ and zero elsewhere, then $$\boxed{f(t)=g(t)+6\mathbf 1_{\{1\}}(t).}$$ Each $u_a$ under our convention is right-continuous. Multiplying it by a continuous function preserves right-continuity, and finite sums preserve that property. Such a sum must equal its right limit at one, so it cannot have the specified value seven while its right limit is one. This is a representational obstruction, not a transform-calculation error.

Step 4: Change the convention without changing the integral. With switch value $1/2$, the corresponding expression for $g$ equals $1/2$ at one and is unchanged elsewhere. Its ordinary transform is still $e^{-s}/(s+1)$. A pointwise correction to recover $f$ would now be $(13/2)\mathbf 1_{\{1\}}$. Thus transform data alone do not specify the value at an isolated discontinuity. Here the right-hand continuous representative is fixed by the chosen convention; an unrestricted pointwise uniqueness claim would be false. The usual uniqueness statement at continuity points is consistent with all these examples.