Prove that $f$ is integrable on $[a,b] $ Suppose that $f(x)=0$ for all $x$ in $[a,b]$ except for some $ z $ in $[a,b]$ Prove that $f$ is integrable on $[a,b] $.
My try:If we can show that $f$ is continuous in $[a,b]$,then the result will follow.Thank you.

Continuity would give integrability, but this function cannot be continuous since $\displaystyle \lim_{x \rightarrow z} f(x) = 0 \neq f(z)$, and this condition must be true for every $y \in [a,b]$ if $f$ is indeed continuous on this interval.  However, it isn't too difficult to tackle the problem directly from the definition:
A function is Riemann integrable on $[a,b] \iff$ for any $\varepsilon > 0$, there exists a partition $ \mathcal{P} = [a\!=\!x_1, \ x_2, \ \cdots, \ x_n\!=\!b]$ of $[a,b]$ such that $U(f, \mathcal{P}) - L(f, \mathcal{P}) < \varepsilon$, where:
$$\displaystyle L(f, \mathcal{P}) = \sum_{i} (x_{i+1} - x_i)\inf \Big( \{f(x) \ | \ x \in [x_i, x_{i+1}] \} \Big)$$ 
$$\displaystyle U(f, \mathcal{P}) = \sum_i (x_{i+1} - x_i)\sup \Big( \{ f(x) \ | \ x \in [x_i, x_{i+1}] \} \Big)$$
and $x_i \in \mathcal{P}$.  Now consider any partition $\mathcal{P}$ of $[a,b]$.  The lower sum is always zero because the infimum of the function values along any interval, even the interval containing $z$, is zero.  Further, the supremum of the function values along any interval is $0$ except for the interval containing $z$.  Therefore:
$$U(f, \mathcal{P}) - L(f, \mathcal{P}) = U(f, \mathcal{P}) = \text{width of the interval containing z}$$
So, is $f$ integrable on $[a,b]$?
