sums and products of measurable functions are measurable

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Real-Valued Functions

Theorem

Let E⊂R be measurable and suppose f,g:E→R are measurable functions, and c∈R. Then cf,f+g, and fg are all measurable functions.

Note

this is useful because we will end up with Lp spaces (L-p spaces) for Lebesgue integrable functions, which are often added together and multiplied.

Proof

(We just check the definition of measurable for each)

cf is measurable

If c=0 then cf=0 is a continuous function and thus measurable

So assume c≠0. If α∈R, then

cf(x)>α⟺f(x)>αc

Thus we have (cf)−1((α,∞])=f−1((αc,∞]) and this is measurable for any α (since f is measurable). Thus cf is measurable.

f+g is measurable

If α∈R, then

f(x)+g(x)>α⟺f(c)>α−g(x)⟺f(x)>r>α−g(x)

Where r is some rational number (since the rationals are dense in the reals +++++). Thus there exists some r∈Q such that

x∈f−1((r,∞])∩g−1((α−r,∞])

And since both f and g are measurable, this intersection is also measurable. Thus the preimage is given as

(f+g)−1((α,∞])=⋃r∈Q(f−1((r,∞]))∩g−1((α−r,∞])

Which is measurable, since this is a countable union. Thus f+g is measurable

fg is measurable

We first show f2 is measurable. Since f2 is a nonnegative function, then for any α<0 we have

(f2)−1((α,∞])=E

because the entire domain maps within (α,∞] and so this is measurable by assumption. If instead α≥0, we have

f2>α⟺f(x)>α or f(x)<−α⟹(f2)−1((α,∞])=f−1((α,∞])∪f−1([−∞,−α))

And both sets on RHS are measurable, so using the equivalent intervals of measurability for measurable functions, we see that the union is measurable and thus f2 is measurable.

Finally, we notice that

fg=14((f+g)2−(f−g)2)
  • And both f+g and f−g are measurable by the previous two proofs above.
◼

Complex-Valued Functions

Theorem

If f,g:E→C are measurable and α∈C, then

Are all measurable as well.

Proof

Exercise

References

References

See Also

Mentions

Mentions

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Created 2025-07-14 Last Modified 2025-07-14