Hi Gowri;

I think the best you can do in approximating such function is Lusin's

Theorem, which is sometimes very useful. Anyways, here I provide a

counter example to your result (which I hope is right).

There exists a sequence {En} of disjoint measurable sets on R such

that

(i) the measure of En is less than 2^n

(ii)there exists a countable basis {In} of the topology of R then En

intersects In in a set of positive measure for every n

(note that you use this construction to solve exercise 8 of Rudin

Chapter 2)

Now consider the function f=sigma{(1/2)^n * (Characteristic function

of En)/ (measure of En)}. Then since the En are disjoint f converges

everywhere. It is easy to see that the L1 norm of f is finite, but its

integral on any interval of R is infinite. This last fact proves that

even if you change f on a set of measure 0, it will remain

discontinuous.

But you tell me: "f may not be in L^infinity". Then take the function

g=exp{-f} then g is bounded and is continuous if and only if f is

continuous (here we needed the fact that f is never infinite, which

explains my choice of this particular f rather than more easier

examples satisfying almost the same properties as f).

g is in L^infinity (bounded). If I can change g on a set of measure 0

to make it continuous then the same is true for f, which is a

contradiction.

Please check the above counter example and keep me in touch if

something is wrong.

Best Regards;

Zaher Hani